exam 2 summary - people · 2 + y b 2 = z2 elliptic cone (cone when a= b= 1) x a 2 + y b 2 = z c 2 +...

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Exam 2 Summary Disclaimer: The exam 2 covers lectures 9-15, inclusive. This is mostly about limits, continu- ity and differentiation of functions of 2 and 3 variables, and some applications. The complete ideas, definitions and theorems are included in the lecture notes and textbook. This is my attempt to help you study, and might not include everything that you need to know. Lecture 9: Functions of Several Variables 1. The domain of a function is the set of all possible inputes of the function and the range is the set of all outputs. 2. The domain of a function of 2 variables is a subset of R 2 and the domain of a function of 3 variables is a subset of R 3 . In either case, the range is a subset of R. 3. The domain restrictions you learnt in Calculus I/ precalculus still apply. These include, but not limited to Domain restriction You must solve for , , 0 something , , 6=0 ln(,) , > 0 tan(,) , 6= (2k - 1)π 2 sin -1 (,), cos -1 (,) -1 , 1 Polynomials, exponential function and inverse tangent have no domain restrictions. 4. To find the range of a given function, freeze all but one variable (at meaningful values) to make your function a function of a single variable and investigate the range. Repeat this with different values/ other variables if necessary. Also keep in mind that the domain restrictions might affect the range. The following table of range of some standard functions from calculus I can be helpful. Function Range x n ( (-∞, ) if n is odd [0, ) if n is even e x (0, ) ln(x), tan(x) (-∞, ) sin(x), cos(x) [-1, 1] tan -1 (x) (-π/2,π/2) sin -1 (x) [-π/2,π/2] cos -1 (x) [0] 1

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Page 1: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +

Exam 2 SummaryDisclaimer: The exam 2 covers lectures 9-15, inclusive. This is mostly about limits, continu-ity and differentiation of functions of 2 and 3 variables, and some applications. The completeideas, definitions and theorems are included in the lecture notes and textbook. This is myattempt to help you study, and might not include everything that you need to know.

Lecture 9: Functions of Several Variables

1. The domain of a function is the set of all possible inputes of the function and the rangeis the set of all outputs.

2. The domain of a function of 2 variables is a subset of R2 and the domain of a functionof 3 variables is a subset of R3. In either case, the range is a subset of R.

3. The domain restrictions you learnt in Calculus I/ precalculus still apply. These include,but not limited to

Domain restriction You must solve for√, , ≥ 0

something

,, 6= 0

ln(,) , > 0

tan(,) , 6= (2k − 1)π

2sin−1(,), cos−1(,) −1 ≤ , ≤ 1

Polynomials, exponential function and inverse tangent have no domain restrictions.

4. To find the range of a given function, freeze all but one variable (at meaningful values) tomake your function a function of a single variable and investigate the range. Repeat thiswith different values/ other variables if necessary. Also keep in mind that the domainrestrictions might affect the range. The following table of range of some standardfunctions from calculus I can be helpful.

Function Range

xn

{(−∞,∞) if n is odd

[0,∞) if n is even

ex (0,∞)ln(x), tan(x) (−∞,∞)sin(x), cos(x) [−1, 1]tan−1(x) (−π/2, π/2)sin−1(x) [−π/2, π/2]cos−1(x) [0, π]

1

Page 2: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +

5. The level curves of a function f(x, y) of 2 variables are the curves you get by settingf(x, y) =Constant. The shape of the curve depends on the constant you choose. Thequestion Characterize all level curves asks you to find all possible different shaped levelcurves obtained by using different constants.

6. The level surfaces of a function f(x, y, z) of 3 variables are the surfaces you get by settingf(x, y, z) =Constant. The shape of the curve depends on the constant you choose. Thequestion Characterize all level surfaces asks you to find all possible different shapedlevel surfaces obtained by using different constants. In lecture 5 you studied somecommonly occurring surfaces. Some of these equations are given below.

Equation Surface(xa

)2+(yb

)2= z2 Elliptic cone (Cone when a = b = 1)(x

a

)2+(yb

)2=(zc

)2+ 1 Hyperboloid (1 sheet)(x

a

)2+(yb

)2=(zc

)2− 1 Hyperboloid (2 sheets)

z = x2 + y2 paraboloidz = x2 − y2 Paraboloid hyperbolic(xa

)2+(yb

)2+(zc

)2= 1 Ellipsoid

Lecture 10: Limits of Multivariable functions

1. We say the limit of the function f(x, y) as (x, y) goes to (a, b) is L if the distancebetween f(x, y) and L can be made arbitrarily small by making the distance between(x, y) and (a, b) sufficiently close.

2. The above definition implies that the limit of the function at (a, b) does not depend onthe path you take to reach the point (a, b). Which also implies that if the calculationof limit along two different paths gives two different values, then the limit at that pointdoes not exist.

3. A function f(x, y) is said to be continuous at a point (a, b) if lim(x,y)→(a,b)

f(x, y) = f(a, b).

This implies

(a) f is defined at the point (a, b)

(b) the limit lim(x,y)→(a,b)

f(x, y) exists.

(c) They have the same numerical value.

If any one of the above conditions fail, then the function is not continuous at the point(a, b).

4. If the function is continuous, you can plug in the value and find the limit.

2

Page 3: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +

5. If you are given a limit question

(a) First plug in and see whether you get a finite value. If it does, then this is thelimit value. If you get an infinite form (like 1/0) then the limit is infinite, andmay not exist.

(b) If you get an indeterminate when you plug in, see whether you can employ a calcI trick like

• Factor and cancel

• Multiply by the conjugate

• Clear the denominator using LCD

to find the limit.

(c) If none of the above works, see whether you can get 2 paths that goes to the givenpoint, and gives you 2 different values when you calculate the limit. In this case,the limit does not exist.

The ideas here can be extended naturally to a function of 3 variables by making the obviousbut significant changes.

Lecture 11: Partial Derivatives of Multivariable func-

tions

1. The partial derivative of a function of two variables f(x, y) with respect to x at thepoint (a, b) is defined by the limit

limx→a

f(x, b)− f(a, b)

x− a:= fx(a, b)

and the partial derivative of f(x, y) with respect to y at (a, b) is the limit

limy→b

f(a, y)− f(a, b)

y − b:= fy(a, b)

Partial derivatives can be denoted in several types of notations. eg:

fx(a, b) =

∣∣∣∣∂f∂x∣∣∣∣(a,b)

= Dxf(a, b)

2. The partial derivative can be computed by thinking of all but the variable in questionto be constants.

3. The partial derivative fx(a, b) gives the rate of change of the function in the directionof x, and fy(a, b) gives the rate of change of f in the direction of y.

4. Higher order partial derivatives can be obtained by successively differentiating thefunction partially with respect to the given order. eg: fxyx2(a, b) means you need topartially differentiate f with respect to x, then with respect to y, and then twice withrespect to x.

3

Page 4: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +

5. Clairaut’s Theorem for mixed partial derivatives: Assume f is defined on an open set Din R2 and both fxy and fyx are continuous through D. Then fxy = fyx for all (x, y) inD.

Lecture 12: Tangent Planes, Linear Approximations and

Differentials.

1. The Linearization of f(a, y) at the point (a, b) is the linear function

L(x, y) = f(a, b) + fx(a, b)(x− a) + fy(a, b)(y − b)

This function can be used to approximate the values of f for points near (a, b).

2. Let f(x, y) 2 variable function f defined in a disk containing (a, b) is and let fx(a, b)and fy(a, b) exist. We say f is differentiable at (a, b) if it can be expressed in the form

f(x, y) = L(x, y) + e(x, y)

Where e(x, y) satisfies

lim(x,y)→(a,b)

e(x, y)√(x− a)2 + (y − b)2

= 0

Note that the existence of partial derivatives at the point (a, b) is not sufficient todetermine whether f is differentiable at (a, b)

3. In this case, the tangent plane to f at the point (a, b) exists and its formula is given by

z = f(a, b) + fx(a, b)(x− a) + fy(a, b)(y − b)

This is the formula for equation of a tangent line of an explicitly defined function.

4. Criterion for differentiability: Assume f has partial derivatives defined in an open set Dcontaining (a, b) with both fx and fy continuous at (a, b). Then f is differentiable at

the point (a, b).

5. If f is differentiable at the point (a, b), then f is continuous at the point (a, b). Theconverse is false.

6. Formulas for differentials for a differentiable function at a point (a, b).

• The difference in independent variables x and y are denoted by dx and dy respec-tively and dx = x− a and dy = y − b.• ∆z is the difference of the functions value between those two points, so ∆z =f(x, y)− f(a, b).

• The differential dz = L(x, y)− f(a, b), where L(x, y) is the linearization functionof f at (a, b). It also gives the equality dz = fx(a, b)(x− a) + fy(a, b)(y − b).• At points (x, y) near (a, b), ∆z ≈ dz so ∆z ≈ fx(a, b)(x− a) + fy(a, b)(y − b).

4

Page 5: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +

Lecture 13: Directional Derivative and the Gradient

Vector

1. The gradient vector of a function f(x, y) of 2 variables is given by

∇f(x, y) = 〈fx(x, y), fy(x, y)〉

For a function g(x, y, z) of 3 variables, this becomes

∇g(x, y, z) = 〈gx(x, y, z), gy(x, y, z), gz(x, y, z)〉

2. Let f be a differentiable function at (a, b) and let u be a unit vector in the xy-plane.The directional derivative of f at the point (a, b) in the direction of u is given by theformula

Duf(a, b) = ∇f(a, b) · u

3. Properties of ∇f .

(a) ∇f is a vector field. That is, every point on f(a, b), which is a surface gets itsown vector ∇f(a, b).

(b) ∇f points in the direction of the maximum increase of f (or the steepest ascent).−∇f points to the direction of the maximum increase of f (or the steepest descent)

(c) Any vector orthogonal to ∇f points to a direction where f does not change.

(d) ∇f is orthogonal to the level curves

∇f has the same properties in 3 variables, and the last property should be changed to:∇f is normal to the level surfaces of f .

Lecture 14: The Chain Rule, Implicit Differentiation,

the Gradient and the level curves

1. If z is a function of x and y, and x and y are in turn functions of an independentvariable t, the chain rule states

dz

dt=∂z

∂x

dx

dt+∂z

∂y

dy

dt

In this case, z is the dependent variable, x and y are intermediate variables and t isthe independent variables.

2. If z is a function of x and y, and x and y are in turn functions of two independentvariables s and t, then the chain rule states

∂z

∂t=∂z

∂x

∂x

∂t+∂z

∂y

∂y

∂tand

∂z

∂s=∂z

∂x

∂x

∂s+∂z

∂y

∂y

∂s

5

Page 6: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +

3. In other cases, you can find the corresponding derivatives/partial derivatives by drawinga variable tree as described in class.

4. Computationally, it is easier to plug in and write the function in terms of the indepen-dent variables and compute the required (partial) derivatives.

5. If F (x, y) = 0 is an implicit function of x and y, then the derivativedy

dxis given by the

formulady

dx= −Fx

Fy

6. As noted in lecture 13, The gradient vector ∇f(x, y) of a 2 variable function is orthog-onal to the level curves, and the gradient vector ∇g(x, y, z) of a 3 variable function isnormal to the level surfaces.

7. Tangent planes for an implicit function F (x, y, z) = 0 at the point (a, b, c) is given by

∇F (a, b, c) · 〈(x− a), (y − b), (z − c)〉 = 0

Lecture 15: Local Maximum and Minimum Values, Sec-

ond Derivative Test, and the Extreme Value Theorem

1. A function f(x, y) has a local maximum value at (a, b) if f(a, b) ≥ f(x, y) for all (x, y)in the domain of f in some open disk containing (a, b). A function f(x, y) has a localminimum value at (a, b) if f(a, b) ≤ f(x, y) for all (x, y) in the domain of f in someopen disk containing (a, b).

2. A point (a, b) in the domain of f is called a critical point if either

(a) Both fx(a, b) = 0 and fy(a, b) = 0 or

(b) One of fx(a, b) or fy(a, b) is not defined.

3. Fermat’s Theorem If f has a local maximum or a minimum at the point (a, b), then(a, b) is a critical number of f .Remark: The converse of Fermat’s theorem is not true. That is, there can be pointswhere the partial derivatives are zero or undefined, yet fail to be a local max or min.eg: Saddle points.Remark: Fermat’s Theorem also implies that if f is differentiable and f has a localmaximum or a local minimum at a point (a, b), then both fx(a, b) = 0 and fy(a, b) = 0.

4. A function f has a saddle point at a critical point (a, b) if, in every open disk centeredat (a, b) there are points (x1, y1) for which f(x1, y1) > f(a, b) and points (x2, y2) forwhich f(x2, y2) < f(a, b).

6

Page 7: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +

5. Second Derivative Test: Suppose that the second order partial derivatives are continuousthroughout an open disk centered at the point (a, b) where fx(a, b) = fy(a, b) = 0. Wecan define a new function D(x, y) called the discriminant by the formula

D(x, y) = fxx(x, y)fyy(x, y)− (fxy(x, y))2

(a) If D(a, b) > 0 and fxx(a, b) < 0, then f has a local maximum at (a, b).

(b) If D(a, b) > 0 and fxx(a, b) > 0, then f has a local minimum at (a, b).

(c) If D(a, b) < 0, then f has a saddle point at (a, b).

(d) If D(a, b) = 0, the second derivative test is inconclusive.

6. A point (a, b) is said to be an absolute maximum of a function f defined in a domain Dif f(a, b) ≥ f(x, y) for all (x, y) ∈ D. A point (a, b) is said to be an absolute minimumif f(a, b) ≤ f(x, y) for all (x, y) ∈ D.

7. Extreme Value Theorem If a function f is continuous on a closed and bounded domainD, then f has both an absolute maximum and an absolute minimum on D.Remark: By Fermat’s Theorem, the absolute maximum or the minimum occurs eitherat critical points or on the boundary.

8. To find absolute extrema in a closed and bounded region

(a) First find partial derivatives to find critical numbers of f on the domain. Evaluatethe function’s value at the critical numbers.

(b) Parametrize the boundary, and find the critical numbers by differentiating. Eval-uate the function on the points you found.

(c) Pick the absolute maximum and the minimum. Make sure they are in the domain.

7

Page 8: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +

L9-L15 Practice problems

1. Find the domain and the range of the following functions

(a) f(x, y) = x2 sin(y)

(b) f(x, y) =ex

y2

(c) f(x, y, z) =sin(z)√

1 + x2 + y2

(d) f(x, y) =sin(y)√1− x2

(e) f(x, y, z) =tan−1(x)

yz

2. Describe the level curves of the functions below.

(a) f(x, y) = x2 − y2

(b) f(x, y) = x2 + y2

(c) f(x, y) =√x+ y.

3. Describe the level surfaces of the following functions

(a) f(x, y, z) = x2 + y2 + 4z2

(b) f(x, y, x) =√

16− x2 − y2 − z2

(c) f(x, y, z) = x2 + y2 − z2

4. Show that the following limit of the function does not exist at the given point by evaluating them in twosuitably chosen paths.

lim(x,y)→(0,0)

x 3√y

x2 + y2/3

5. Evaluate the limit using continuity

(a) lim(x,y)→(1,1)

ex2 − e−y2

x+ y

(b) lim(x,y)→(2,3)

tan−1(x2 − y)

6. Evaluate the following limit

lim(x,y)→(0,0)

6xy√x2 + y2

or show that it does not exist.

Page 9: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +

7. Evaluate the following limits or show that it does not exist.

(a) lim(x,y)→(0,0)

y − 2√x2 − 4

(b) lim(x,y)→(0,0)

xy√x2 + y2

(c) lim(x,y)→(0,0)

(x+ y + 2)e−1/(x2+y2)

2

(d) lim(x,y)→(0,0)

tan

(π − (x2 + y2)

4

)tan−1

(1

x2 + y2

)(e) lim

(x,y)→(0,0)

x2 + y2√x2 + y2 + 1− 1

(f) lim(x,y)→(0,0)

(1 + x)y/x

8. Compute all the first order partial derivatives of the following functions.

(a) z =√

4− x2 − y2

(b) z = (sin(x))(sin(y))

(c) z = sin(u2v)

(d) f(x, y, z) =x

y + z

(e) f(x, y) = xy

9. Find the equation of the tangent plane for the following functions at the given points.

(a) f(x, y) = x2y + xy3 at (2, 1)

(b) f(x, y) =x√y

at (4, 4)

(c) g(x, y) = ex/y at (2, 1)

10. Find points on z = xy3 + 8y−1 where the tangent plane is parallel to 2x+ 7y + 2z = 0.

11. Write the linear approximation to f(x, y) = x(1 + y)−1 at (a, b) = (8, 1). Then use it to estimate7.9

2.1

12. Let I =W

H2be the Body Mass Index (BMI). A boy has W = 65kg and height H = 1.5m. Measure the

change in I if (W,H) changes to (66, 1.55)

13. Calculate the directional derivative of the following functions in the direction of the vector u at the givenpoint.

(a) f(x, y) = x2y3 in the direction of u = ı+ 2 at the point (−2, 1).

(b) f(x, y) = sin(x− y) in the direction of u = 〈1, 1〉 at the point (π/2, π/6).

(c) f(x, y) = exy−y2

in the direction of u = 〈12,−5〉 at the point (2, 2).

(d) g(x, y, z) = xe−yz in the direction of u = 〈1, 1, 1〉 at the point (1, 2, 0).

14. Find the directional derivative of f(x, y) = x2 + 4y2 at the point P = (3, 2) in the direction pointing to theorigin.

15. Suppose that for a point P , ∇fp = 〈2,−4, 4〉. Is f increasing or decreasing in the direction of u = (2, 1, 3)

2

Page 10: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +

16. (a) Evaluate ∇f if f(x, y) = xy2.

(b) What is ∇f(2, 3)?

(c) Find the unit vector that points in the direction of maximum change at (2, 3).

17. Let f(x, y) = tan−1(x

y

)and u =

⟨1√2,

1√2

⟩.

(a) Calculate the gradient of f

(b) Calculate Duf(1, 1) and Duf(√

3, 1).

(c) Show that the lines y = mx for m 6= 0 are level curves of f .

18. Use the chain rule to calculate the partial derivatives, express the answer in terms of the independentvariable.

(a) f(x, y, z) = xy + z2. Where x = s2, y = 2rs and z = r2.

Find∂f

∂sand

∂f

∂r.

(b) R(x, y) = (3x+ 4y)5. Where x = u2 and y = uv. Find∂R

∂uand

∂R

∂v.

(c) f(x, y, z) = xy − z2. Where x = r cos(θ), y = cos2(θ) and z = r. Find∂f

∂θ.

(d) f(x, y, z) = x3 + yz3 where x = u2v, y = u+ v2 and z = uv. Find fu(−1,−1) and fv(−1,−1).

19. Let let z be given as an implicit function F (x, y, z) = x2z + y2z + xy − 1 = 0.

(a) Find Fx, Fy and Fz.

(b) Using (a), find∂z

∂xand

∂z

∂y.

(c) Find the equation of the tangent plane of the above given surface at the point (1, 0, 1).

20. Find the equation of the equation of the tangent line to the implicitly defined function x2 + y2 − z2 = 1 atthe point (1, 1, 1).

21. Find all the critical points off(x, y) = 8y4 + x2 + xy − 3y2 − y3

22. (a) Find the critical numbers of the function f(x, y) = y2x− yx2 + xy.

(b) Use the second derivative test to determine the nature of the critical points.

23. For the following functions, first find all the critical numbers, and then determine the nature of the criticalpoints using the second derivative test.

(a) f(x, y) = x2 + y2 − xy + x.

(b) f(x, y) = 4x− 3x2 − 2xy2

(c) f(x, y) = x3 − xy + y3

(d) f(x, y) = ex − xey

24. Find the absolute extrema of the following functions on the given domain

(a) f(x, y) = (x2 + y2 + 1)−1 on 0 ≤ x ≤ 3 and 0 ≤ y ≤ 5.

(b) f(x, y) = e−x2−y2

, x2 + y2 ≤ 1.

(c) f(x, y) = x+ y − x2 − y2 − xy on 0 ≤ x ≤ 2 and 0 ≤ y ≤ 2.

25. Find the point on the plane x− x− y = 1 closest to the point (1, 0, 0).

3

Page 11: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +

26. Find the maximum volume of the largest box of the type shown in the following figure, with one corner atthe origin and the opposite cornere at a point P = (x, y, z) on the paraboloid

z = 1− x2

4− y2

9

with x, y, z ≥ 0.

4

Page 12: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +

MAC2313 Test 2A

(5 pts) 1. The equation of the tangent plane to surface 25x2 + 4y2 + 4z2 = 100 at the point(0, 4, 3) is:

A. 32y + 24z = 200 B. 8x+ 24y + 32z = 200 C. 8x+ 32y + 24z = 200

D. 24y + 32z = 200 E. none of the above

(5 pts) 2. If w is a function of the variables x, y, and z and each of the variables x, y, and z isa function of the variables s and t, then which of the following is true?

A.∂w

∂s=

∂w

∂x

∂x

∂s+

∂w

∂y

∂y

∂s+

∂w

∂z

∂z

∂sB.

dw

ds=

∂w

∂x

dx

ds+

∂w

∂y

dy

ds+

∂w

∂z

dz

ds

C.dw

ds=

∂w

∂x

∂x

∂s+

∂w

∂y

∂y

∂s+

∂w

∂z

∂z

∂sB.

∂w

∂s=

∂w

∂x

dx

ds+

∂w

∂y

dy

ds+

∂w

∂z

dz

ds

E. none of the above

(5 pts) 3. The derivative∂2

∂x∂y (y + y2ex) is equal to:

A. yex B. 2yex C. x+ y + yex D. 2x+ 2y + 2yex E. 1 + 2yex

(5 pts) 4. If u =⟨2/√13, 3/

√13

⟩and f(x, y) = x2 + y, then Duf(1, 0) is equal to:

A. 2/√13 B. 3/

√13 C. 5/

√13 D. 6/

√13 E. 7/

√13

Page 13: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +

(5 pts) 5. Which of the following are critical points of the function f(x, y) = x2+2y2−4xy−6x?

I. (3, 3)

II. (−3,−3)

III. (0, 0)

IV. (1, 1)

A. only I B. only II C. only III and IV D. only I and II E. only I, III, and IV

(5 pts) 6. The equation of the tangent plane to the surface z = cos(xy) at the point (2, π, 1) isgiven by:

A. L(x, y) = πx+ 2y B. L(x, y) = πx+ 2y − 2π C. L(x, y) = πx+ 2y + 2π

D. L(x, y) = πx+ 2y + 4π E. none of the above

(5 pts) 7. Given the function g(x, y, z) =√16− x2 − y2 − z2, which of the following are true?

I. D = { (x, y, z) | 16 ≥ x2 + y2 + z2 }

II. R = [0, 16]

III. The level surfaces are spheres.

IV. The graph of the function g(x, y, z) lies in R4.

A. only I B. only III C. only II and IV D. only I and II E. only I, III, and IV

Page 14: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +

(5 pts) 8. If f(x, y) = (2xy + 3y2)3, then fx(−1, 1) is equal to:

A. −4 B. 4 C. 6 D. 8 E. 12

(5 pts) 9. Let (a, b) be a point in the domain of the real-valued function f(x, y) and let L ∈ R;which of the following are true?

I. If f(x, y) is continuous at (a, b), then lim(x,y)→(a,b) f(x, y) exists.

II. If lim(x,y)→(a,b) f(x, y) exists, then f(x, y) is continuous at (a, b).

III. If f(x, y) is continuous at (a, b), then lim(x,y)→(a,b) f(x, y) = f(a, b).

IV. If lim(x,y)→(a,b) f(x, y) = L, then lim(x,y)→(a,b)

√f(x, y) =

√L.

A. only I B. only II C. only I and II D. only II and IV E. I and III

(5 pts) 10. For f(x, y) = x3 + y3 − 12xy, the critical point (4, 4) is:

A. a local max B. a local min C. a saddle point D. a boundary point

E. the second derivative test fails at this point

(5 pts) 11. To find the point on the surface z = x2 − y2 closest to the point (1, 2, 3), which ofthe following functions should be optimized?

A.√x2 + y2 + z2 B.

√(x+ 1)2 + (y + 2)2 + (z + 3)2 C.

√x2 + y2 + (x2 − y2)2

D. (x− 1)2 + (y − 2)2 + (x2 − y2 − 3)2 E. none of the above

Page 15: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +

(5 pts) 12. Which of the following is a vector orthogonal to the tangent line to the curvex2 − 4xy − y2 = 4 at the point (5, 1)?

A. ⟨−6, 2⟩ B. ⟨1, 3⟩ C. ⟨−1, 3⟩ D. ⟨6,−22⟩ E. ⟨−11, 3⟩

(5 pts) 13. Let f(x, y, z) = x2 + yz + cos(πz), x = t, y = cos t, and z = et. Calculate the valueof derivative of the f(x, y, z) with respect to t when t = 0.

A. 0 B. 1 C. −1 D. 1 + π E. 1− π

(5 pts) 14. Bonus. Let f(x, y) function which is differentiable at the point (a, b) and let u be anonzero vector in R2. Which of the following are true?

I. The equation of the tangent plane to the surface at (a, b, f(a, b)) is given byL(x, y) = fy(a, b)(x− a) + fx(a, b)(y − b).

II. Duf(a, b) = ⟨fx(a, b), fy(a, b)⟩ · u

III. The max rate of change of f(x, y) at (a, b) is in the direction of ∇f(a, b).

IV. If ∇f(a, b) = 0, then (a, b) is a critical point of f(x, y).

A. I and III B. II and III C. I and IV D. III and IV E. II and IV

Page 16: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +

MAC2313 Test 2A

Name: UF-ID: Section:

(6 pts) 1. Calculate the following limit:

lim(x,y)→(2,2)

x2 + y2

x− y.

(6 pts) 2. Given f(x, y) = 2x2 + y2, use the value of the function at the point (2, 1) anddifferentials to approximate f(1.95, 1.1).

Page 17: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +

(7 pts) 3. Use the technique of implicit differentiation discussed in class to calculatedydx given√

x2y2 + 3y4 = 2e3.

(6 pts) 4. Given f(x, y) = x sin y + x2, at the point (1, π) find a vector which points in thedirection of no rate of change of the function.

Page 18: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +

(10 pts) 5. Given f(x, y) = (x2−1)(y2−4), fx(x, y) = 2x(y2−4), and fy(x, y) = 2y(x2−1), findthe max and min values of f(x, y) on the rectangular region bounded by the lines x = 1, x = −1,y = 2, and y = −2.

Page 19: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +

MAC2313 Test 2 A

(5 pts) 1. The equation of the tangent plane to the surface z = 2x2 − 3xy + 4y2 at (1, 2, 12) isgiven by:

A. 2x−13y+z = −12 B. 2x−13y+z = −24 C. −2x+13y+z = 12 D. −2x+13y+z = 24

E. none of the above

(5 pts) 2. Given a function z = f(x, y) with continuous first partial derivatives at (a, b), whichof the following are true for (x, y) close to (a, b)?

I. f(x, y) ≈ fx(a, b)(x− a) + fy(a, b)(y − b)

II. ∆z ≈ fx(a, b)(x− a) + fy(a, b)(y − b)

III. f(x, y) ≈ dz + f(a, b)

A. only I B. only II C. only III D. only I and III E. only II and III

(5 pts) 3. If g(x, y, z) = xyz2 + cos(zy) + e2x−3y, what is the value of the derivative∂3g

∂y2∂zat

the point (1, 0, 2)?

A. −4 B. −2 C. 0 D. 2 E. 4

(5 pts) 4. What is the domain of the function f(x, y) =−x ln y

1−x2−y2?

A. { (x, y) | y ≥ 1, x2 + y2 < 1 } B. { (x, y) | y ≥ 1, x2 + y2 > 1 }

C. { (x, y) | y ≥ 0, x2 + y2 = 1 } D. { (x, y) | y > 0, x2 + y2 < 1 }

E. { (x, y) | y > 0, x2 + y2 = 1 }

Page 20: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +

(5 pts) 5. What is the range of the function f(x, y) =ln(x2+y2)2+sin(x+y)?

A. (−∞,∞) B. (0,∞) C. [0,∞) D. (−∞, 0) E. (−∞, 0]

(5 pts) 6. The level curves of z2 = 4x2 + 4y2, |z| > 0 are all circles.

A. true B. false

(5 pts) 7. A level curve is a trace generated by intersecting the graph of a function with a planeparallel to the x, y-coordinate plane.

A. true B. false

(5 pts) 8. Which of the following are true?:

I. along the curve y = 0, lim(x,y)→(0,0) (y2 − x)/(y + x) = −1

II. along the curve x = 0, lim(x,y)→(0,0) (y2 − x)/(y + x) = 0

III. along the curve y = x, lim(x,y)→(0,0) (y2 − x)/(y + x) = −1/2

IV. the lim(x,y)→(0,0) (y2 − x)/(y + x) exists

A. only I B. only II C. only III D. only I, II, and III E. I, II, III and IV

Page 21: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +

(5 pts) 9. Given a function f(x, y) such that ∇f(a, b) = 0 and f(a, b) = z0, ∇f(a, b) is parallelto the tangent line of the level curve f(x, y) = z0 at (a, b).

A. true B. false

(5 pts) 10. Let f(x, y) be a function differentiable at the point (a, b) and let v = ⟨h, k⟩ be a unitvector; which of the following are true?

I. Dvf(a, b) > 0

II. Dvf(a, b) = ∇f(a, b) · v

III. the vector v points in the direction of the maximum rate of change of f at (a, b)

IV. Dvf(a, b) = |∇f(a, b)| sin θ where θ is the angle between v and ∇f(a, b)

A. only I B. only II C. only II and III D. only II and IV E. I, II, III, and IV

(5 pts) 11. Given g(x, y, z) = 2x2 − 3yz3, ∇g(1, 0,−1) is orthogonal to which of the followingvectors:

A. ⟨3, 1,−2⟩ B. ⟨0, 2, 1⟩ C ⟨0,−2, 1⟩ D. ⟨0, 2, 0⟩ E. ⟨−3, 4, 7⟩

(5 pts) 12. If the Method of Lagrange Multipliers is used to find the maximum value off(x, y, z) = 2x2 + 3xy2 + 4z3 on the surface x2 + 2y2 + 5z2 = 10, which of the following is notone of the equations which must be solved simultaneously:

A. 12z2 = 10λz B. x2 + 2y2 + 5z2 = 10 C. 4x+ 6xy = 2λx D. 6xy = 4λy

Page 22: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +

(5 pts) 13. Which of the following vectors point in the direction of the maximum rate of decreaseof f(x, y) = e2x+3y cos y at the point (0, 0):

A. ⟨3, 2⟩ B. ⟨2, 3⟩ C ⟨−2,−3⟩ D. ⟨−3, 2⟩ E. none of the above

(5 pts) 14. Bonus. Given f(x, y) = 2x2 + 3y − 4, if the linearization of this function at (0, 0) isused to approximate the function at (0.1,−0.2), what value is obtained?

A. −4 B. −3.8 C. −3.6 D. −4.6 E. none of the above

(5 pts) 15. Bonus. Let f(x, y) be a function with continuous second partial derivatives and let(a, b) be a point such that fx(a, b) = fy(a, b) = 0; using the second derivative test, which of thefollowing are true?

I If D(a, b) < 0 and fxx(a, b) > 0, then f has a local max value at (a, b).

II. If D(a, b) = 0, then f has a saddle point at (a, b).

III. If D(a, b) < 0 and fxx(a, b) < 0, then f has a local min value at (a, b).

IV. D(a, b) = fxx(a, b)fyy(a, b)− [fxy(a, b)]2.

A. only II B. only IV C. only I and III D. only I, III, and IV E. only I, II and III

Page 23: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +

MAC2313 Test 2A

Name: UF-ID: Section:

(7 pts) 1. If w = 2xz − ey, x = s + t, y = s − 3t, and z = st, write the chain rule for thederivative of w with respect to t; use the chain rule to calculate the derivative.

(5 pts) 2. Find the critical point(s) of the function f(x, y) = x2(xy + 12)− 8y.

Page 24: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +

(6 pts) 3. Find the vector equation of the tangent plane to the ellipsoid x2 + 3y2 + 4z2 = 8 atthe point (1,−1, 1).

(7 pts) 4. Characterize the critical point (1,−1) of the function f(x, y) = x+ x2y + y + xy2.

Page 25: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +

(10 pts) 5. Use the Method of Lagrange Multipliers to find the maximum and minimum valuesof f(x, y) = 2x− 6y − 1 on the ellipse x2 + 3y2 = 1.

Page 26: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +

MAC2313 Test 2A

(5 pts) 1. Let z = x2 + ln y. How many of the following are true?

i. The domain of the function is { (x, y) | y ≥ 0 }.

ii. The range of the function is [0,+∞).

iii. The given function is explicit.

iv. The graph of the function is a surface in R3.

A. 0 B. 1 C. 2 D. 3 E. 4

(5 pts) 2. A level curve is a trace generated by intersecting a surface with a plane parallel tothe y, z-coordinate plane.

A. True B. False

(5 pts) 3. Let f(x, y) =x2−yx2+y

. How many of the following are true?

i. Along the curve y = x2, lim(x,y)→(0,0) f(x, y) = 0.

ii. Along the line x = 0, lim(x,y)→(0,0) f(x, y) = 1.

iii. Along the line y = 0, lim(x,y)→(0,0) f(x, y) = −1.

iv. The limit, lim(x,y)→(0,0) f(x, y), exists.

A. 0 B. 1 C. 2 D. 3 E. 4

Page 27: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +

(5 pts) 4. If g(x, y, z) = z cos(x2z−y2) what is the value of∂2g∂x∂y when (x, y, z) = (

√2π,

√π, 1)?

A. 0 B. −8π C. −4π D. −4√2π E. none of the above

(5 pts) 5. The equation of the plane tangent to z = 3x2 + 4y2 when (x, y) = (1, 3) is given by:

A. z = 8x+ 18y− 35 B. z = 4x+ 24y− 39 C. z = 6x+ 24y + 39 D. z = 8x+ 18y + 35

E. none of the above

(5 pts) 6. Let z = 3x3y; what is the approximate value of the change of z when (x, y) changesfrom (2, 1) to (1.9, 1.05)?

A. −0.64 B. −2.4 C. 0.46 D. 1.4 E. 2.8

Page 28: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +

(5 pts) 7. If f(x, y) = 5y sin x + y5, what is Duf(0, 1) when u points in the direction of thevector ⟨3, 4⟩?

A. −√5 B. 0 C.

√5 D. 5 E. none of the above

(5 pts) 8. Consider the function f(x, y) = −2ex2+3y at the point (x, y) = (0, 0); which of the

following vectors gives the direction in which the function is not changing?

A. ⟨0, 6⟩ B. ⟨6, 0⟩ C. ⟨3, 4⟩ D. ⟨3, 1⟩ E. ⟨0,−6⟩

(5 pts) 9. Given a function f(x, y) differentiable at a point (a, b), the line tangent to the levelcurve of f at (a, b) is orthogonal to the gradient ∇f(a, b) provided ∇f(a, b) = 0.

A. True B. False

(5 pts) 10. The function f(x, y) = 2y + x2y − 3xy has how many critical points?

A. 0 B. 1 C. 2 D. 3 E. 4

Page 29: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +

(5 pts) 11. For f(x, y) = x4 + y4 − 4xy + 1, the critical point (0, 0) is:

A. a local min B. a local max C. a saddle point D. a boundary point

E. the second derivative test fails at this point

(5 pts) 12. If we use the Method of Lagrange Multipliers to determine the max value off(x, y, z) = 5x − 3y + z on the ellipsoid x2 + y2 + 4z2 = 10, how many of the following equa-tions are included in the system which must be simultaneously solved?

i. 1 = 8λz

ii. x2 + y2 + 4z2 = 0

iii. 3 = 2λy

iv. 1 = 4λz

A. 0 B. 1 C. 2 D. 3 E. 4

Page 30: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +

(5 pts) 13. Let f(x, y) be a twice differentiable function on a region D of the x, y-plane and let(a, b) be an interior point of D. How many of the following are true?

i. The function is continuous at (a, b).

ii. The discriminant of f(x, y) at (a, b) is equal to fxx(a, b)fyy(a, b)− fxy(a, b).

iii. It is possible to draw a sufficiently small circle surrounding (a, b) such that every point inthe circle is in D.

iv. The point (a, b) is a critical point of f(x, y).

A. 0 B. 1 C. 2 D. 3 E. 4

(5 pts) 14. Bonus. The level surface 9x2 + 7y2 + 2z2 = C, is defined for all real numbers C.

A. True B. False

Page 31: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +

MAC2313 Test 2A

Name: UF-ID: Section:

(5 pts) 1. The function f is a function of the single variable t and in turn, t is a function of thevariables u, v, and w; using the chain rule, give an expression for the derivative of the function fwith respect to w.

(8 pts) 2. The density of a thin circular plate of radius 3 is given by ρ(x, y) = 6 + xy and theedge of the plate is described by the parametric equations x = 3 cos t, y = 3 sin t, for 0 ≤ t ≤ 2π.Use the chain rule to find the rate of change of density with respect to t on the edge of the plate.Express your final answer in terms of t.

Page 32: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +

(10 pts) 3. Find and classify all of the critical points of the function f(x, y) = x2+6xy+3y2−4y.

Page 33: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +

(12 pts) 4. Use the Method of Lagrange Multipliers to find the min value off(x, y, z) = 2x2 + y2 + 2z2 on the plane 2x+ 3y + 4z = −19.

Page 34: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +
Page 35: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +
Page 36: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +
Page 37: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +
Page 38: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +
Page 39: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +
Page 40: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +
Page 41: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +
Page 42: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +
Page 43: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +
Page 44: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +
Page 45: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +
Page 46: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +
Page 47: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +
Page 48: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +
Page 49: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +
Page 50: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +
Page 51: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +
Page 52: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +
Page 53: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +
Page 54: Exam 2 Summary - People · 2 + y b 2 = z2 Elliptic cone (Cone when a= b= 1) x a 2 + y b 2 = z c 2 + 1 Hyperboloid (1 sheet) x a 2 + y b 2 = z c 2 1 Hyperboloid (2 sheets) z= x2 +