daily webwork, #1

Daily WeBWorK, #1

Consider the ellipsoid x2 + 3y2 + z2 = 11. Find all the points where thetangent plane to this ellipsoid is parallel to the plane 2x + 3y + 2z = 0.

I In order for the plane tangent to the ellipsoid at a point (a, b, c) to beparallel to the plane 2x + 3y + 2z = 0, the two planes’ normal vectorsmust be parallel.

I This means the two planes’ normal vectors must be multiples of eachother.

I A vector normal to the plane 2x + 3y + 2z = 0 is < 2, 3, 2 >.

I Goal: Find all points (a, b, c) where a vector normal to the tangent tothe ellipsoid at the point (a, b, c) is k< 2, 3, 2 > for some value of k.

Math 236-Multi (Sklensky) In-Class Work April 5, 2013 1 / 32

Daily WeBWorK, #1, continuedConsider the ellipsoid x2 + 3y2 + z2 = 11. Find all the points where thetangent plane to this ellipsoid is parallel to the plane 2x + 3y + 2z = 0.

Goal: Find all points (a, b, c) where a vector normal to the tangent to theellipsoid at the point (a, b, c) is k< 2, 3, 2 > for some value of k .

To find a vector normal to the ellipsoid at the point (a, b, c):

I Define the function f (x , y , z) = x2 + 3y2 + z2 − 11.

I Our specific ellipsoid is the level surface for this function withf (x , y , z) = 0.

I Showed last time: gradients are normal to level surfaces.

Thus the gradient−→∇f (a, b, c) will be normal to the tangent plane to

the ellipsoid at the point (a, b, c).

I−→∇f (x , y , z) =< 2x , 6y , 2z >

I Thus at a point (a, b, c), < 2a, 6b, 2c > is normal to the tangentplane to the ellipsoid.Math 236-Multi (Sklensky) In-Class Work April 5, 2013 2 / 32



I At a point (a, b, c), < 2a, 6b, 2c > is normal to the tangent plane tothe ellipsoid.

I We want to find all points (a, b, c) where< 2a, 6b, 2c >= k < 2, 3, 2 > for some k .

2a = 2k ⇒ a = k

6b = 3k ⇒ b = k/2

2c = 2k ⇒ c = k

Thus all points (k , k/2, k) which also lie on the ellipsoid will havetheir tangent planes parallel to the given plane.




I At a point (a, b, c), < 2a, 6b, 2c > is normal to the tangent plane tothe ellipsoid.

I All points (k , k/2, k) which also lie on the ellipsoid will have theirtangent planes parallel to the given plane.

I Find all values of k for which (k)2 + 3

(k2

)2

+ (k)2 = 11.

I Simplifying and solving for k , we find that k2 = 4, so k = ±2

I Thus at (2, 1, 2) and at (−2,−1,−2), the plane tangent to theellipsoid is parallel to 2x + 3y + 2z = 0.


Wrapping up the gradient:As your book says:

always think of gradients as vector-valuedfunctions whose values

I specify the direction of maximum increaseof a function,

and

I provide normal vectors – to the levelcurves, in two dimensions, and to the levelsurfaces in three dimensions.

I Remember that this also gives us a wayto find planes tangent to surfaces


Recall: Critical points for a function f (x , y):

If f (x , y) is a smooth function, so the partials fx(x , y) and fy (x , y) existeverywhere, then the point

(a, b, f (a, b)

)can only be a local maximum,

local minimum, or saddle point if

I f (x , y) is flat at(a, b, f (a, b)

)in all directions

I that is, only if D−→u f (a, b) = 0 for all unit vectors −→u ∈ R2

Thus if fx(a, b) 6= 0 or fy (a, b) 6= 0, the point(a, b, f (a, b)

)can not be a

local maximum, local minimum, or saddle point.

We therefore call all inputs (a, b) such that−→∇f (a, b) =

−→0 critical points.

Note: This does not mean that every critical point is a local maximum,local minimum, or saddle point.


Recall from Calc 1:

For g(x), if g ′(a) = 0 (so that g is flat at x = a), then the 2nd DerivativeTest gives

g ′(a) > 0 =⇒ g is concave up at x = a =⇒ x = a is a minimum

g ′(a) < 0 =⇒ g is concave down at x = a =⇒ x = a is a maximum

g ′(a) = 0 =⇒ the test is inconclusive

Is there an analogous result for f (x , y)?


Question:

Suppose f (x , y) is a smooth function and

I fx(x0, y0) = fy (x0, y0) = 0

I fxx(x0, y0) > 0, fyy (x0, y0) > 0 – That is, in both the x and ydirections, f is concave up

Must f have a local min at the point (x0, y0)?

No!


Example:Let f (x , y) = x2 + 3xy + 2y2 − 9x − 11y .

fx(x , y) = 2x + 3y − 9 =⇒ fx(−3, 5) = −6 + 15− 9 = 0

fy (x , y) = 3x + 4y − 11 =⇒ fy (−3, 5) = −9 + 20− 11 = 0

fxx(x , y) = 2 =⇒ fxx(−3, 5) > 0 fyy (x , y) = 4 =⇒ fyy (−3, 5) > 0

Thus at (−3, 5), f is flat in both the x and y directions, and is concave upin both the x and y directions ...

but f does not have a minimum there:


Example:Let f (x , y) = x2 + 3xy + 2y2 − 9x − 11y .

fx(x , y) = 2x + 3y − 9 =⇒ fx(−3, 5) = −6 + 15− 9 = 0

fy (x , y) = 3x + 4y − 11 =⇒ fy (−3, 5) = −9 + 20− 11 = 0

fxx(x , y) = 2 =⇒ fxx(−3, 5) > 0 fyy (x , y) = 4 =⇒ fyy (−3, 5) > 0

Thus at (−3, 5), f is flat in both the x and y directions, and is concave upin both the x and y directions ... but f does not have a minimum there:


Recall:

As long as fxy (x , y) and fyx(x , y) are continuous, they are equal:

fxy (x , y) = fyx(x , y)


In Class Work

1. Locate all critical points and classify them using the 2nd DerivativesTest:

(a) f (x , y) = x3 − 3xy + y3

(b) f (x , y) = y2 + x2y + x2 − 2y

2. Let f (x , y) = 4xy − x3 − 2y2

(a) Find and classify all critical points of f .

(b) Find the maximum value of f (x , y) on the unit disk{(x , y) | x2 + y2 ≤ 1}.


Solutions

1. Locate all critical points and classify them using the 2nd DerivativesTest:(a) f (x , y) = x3 − 3xy + y3

∂f

∂x= 0 =⇒ 3x2 − 3y = 0 =⇒ y = x2

∂f

∂y= 0 =⇒ −3x + 3y2 = 0 =⇒ x = y2

Putting those two results together,

y = x2 = (y2)2 =⇒ y(y3 − 1) = 0 =⇒ y = 0, y = 1

Thus our critical points are (0, 0) and (1, 1).


Solutions


I We found the only critical points are (0, 0) and (1, 1)

I To classify them, we will use the Second Derivatives Test.

To use that, we need to find fxx(a, b) and the discriminant D(a, b):

fxx(x , y) =∂

∂x(3x2 − 3y) = 6x

D(x , y) =

∣∣∣∣fxx(x , y) fxy (x , y)fxy (x , y) fyy (x , y)

∣∣∣∣ =

∣∣∣∣6x −3−3 6y

∣∣∣∣ = 36xy − 9


Solutions



I Also found fxx(x , y) = 6x and D(x , y) = 36xy − 9

I At (0, 0),

fxx(0, 0) = (6)(0) = 0 and D(0, 0) = (36)(0)(0)− 9 < 0.

According to the Second Derivatives Test, because D < 0, f has asaddle point at the point (0, 0).


Solutions



I We found (0, 0) is a saddle point.

I fxx(x , y) = 6x and D(x , y) = 36xy − 9

I At (1, 1),

fxx(1, 1) = (6)(1) > 0 and D(1, 1) = (36)(1)(1)− 9 > 0.

According to the Second Derivatives Test, because D > 0 andfxx > 0, f has a local minimum at (1, 1).


Solutions1. Locate all critical points and classify them using the 2nd DerivativesTest:(a) f (x , y) = x3 − 3xy + y3


I We found (0, 0) is a saddle point and (1, 1) is a local minimum.

We can see these results on a contour plot of f (x , y):


Solutions

1. Locate all critical points and classify them using the 2nd DerivativesTest:(b) f (x , y) = y2 + x2y + x2 − 2y

fx = 0 =⇒ 2xy + 2x = 0 =⇒ 2x(y + 1) = 0 =⇒ x = 0 or y = −1

fy = 0 =⇒ 2y + x2 − 2 = 0

=⇒ If x = 0, 2y + 02 − 2 = 0 =⇒ y = 1

=⇒ If y = −1, 2(−1) + x2 − 2 = 0 =⇒ x2 = 4 =⇒ x = ±2

Thus our critical points are (0, 1), (2,-1), (−2,−1).


Solutions


I Our critical points are (0, 1), (2,-1), (−2,−1).

I To classify them using the 2nd Derivatives Test, we need:

fxx(x , y) =∂

∂x(2xy + 2x) = 2y + 2

D(x , y) =

∣∣∣∣fxx(x , y) fxy (x , y)fxy (x , y) fyy (x , y)

∣∣∣∣ =

∣∣∣∣2y + 2 2x2x 2

∣∣∣∣ = 4y + 4− 4x2


Solutions



I fxx(x , y) = 2y + 2 and D(x , y) = 4y + 4− 4x2

I At (0, 1)

fxx(0, 1) = 2(1) + 2 > 0 and D(0, 1) = 4(1) + 4− 4(0) > 0.

According to the 2nd Derivatives Test, because D > 0 and fxx > 0, fhas a local minimum at (0, 1).


Solutions



I (0, 1) is a local minimum.


I At (2,−1)

fxx(2,−1) = 2(−1)+2 = 0 and D(2,−1) = 4(−1)+4−4(2)2 < 0.

According to the 2nd Derivatives Test, because D < 0, f has a saddlepoint at (2,−1).


Solutions



I (0, 1) is a local minimum and (2,−1) is a saddle point.


I At (−2,−1)

fxx(−2,−1) = 2(−1)+2 = 0 and D(2,−1) = 4(−1)+4−4(−2)2 < 0.

According to the 2nd Derivatives Test, because D < 0, f has a saddlepoint at (−2,−1).


Solutions1. Locate all critical points and classify them using the 2nd DerivativesTest:(b) f (x , y) = y2 + x2y + x2 − 2y

I Our critical points are (0, 1), (2,-1), (−2,−1).I (0, 1) is a local minimum and both of (±2,−1) are saddle points.

We can again see these results illustrated:


Solutions

2. Let f (x , y) = 4xy − x3 − 2y2


∂f

∂x= 0 =⇒ 4y − 3x2 = 0 =⇒ y =

3

4x2

∂f

∂y= 0 =⇒ 4x − 4y = 0 =⇒ y = x

x = y =3

4x2 =⇒ x

(1− 3

4x

)= 0 =⇒ x = 0 or x =

4

3.

x = 0 =⇒ y = 0 x =4

3=⇒ y =

4

3.

Thus our critical points are (0, 0) and (4/3, 4/3).


Solutions

2. Let f (x , y) = 4xy − x3 − 2y2


I The critical points are (0, 0) and (4/3, 4/3).

I To classify them using the 2nd Derivatives Test, we need:

fxx(x , y) =∂

∂x(4y − 3x2) = −6x

D(x , y) =

∣∣∣∣fxx fxyfxy fyy

∣∣∣∣ =

∣∣∣∣−6x 44 −4

∣∣∣∣ = 24x − 16


Solutions

2. Let f (x , y) = 4xy − x3 − 2y2



I fxx(x , y) = −6x and D(x , y) = 24x − 16

I At (0, 0),

fxx(0, 0) = (−6)(0) and D(0, 0) = 24(0)− 16 < 0.

By the 2nd Derivatives Test, because D < 0, f has a saddle point at(0, 0).


Solutions

2. Let f (x , y) = 4xy − x3 − 2y2



I fxx(x , y) = −6x and D(x , y) = 24x − 16

I f has a saddle point at (0, 0).

I At (4/3, 4/3),

fxx(4/3, 4/3) = (−6)(4/3) < 0 and D(4/3, 4/3) = 24(4/3)−16 > 0.

By the 2nd Derivatives Test, because D > 0 and fxx < 0, f has alocal maximum.


Solutions2. Let f (x , y) = 4xy − x3 − 2y2

(a) Find and classify all critical points of f .I The critical points are (0, 0) and (4/3, 4/3).I fxx(x , y) = −6x and D(x , y) = 24x − 16I f has a saddle point at (0, 0).I f has local maximum at (4/3, 4/3).

To illustrate these points:



(b) Find the max value of f (x , y) on the unit disk {(x , y) | x2 + y2 ≤ 1}.We know that f has a saddle point at (0, 0) and a local maximum at(4/3, 4/3), and that these are the only critical points.

We can see that the maximum value will occur somewhere on theboundary of the unit disk.


Solutions

2. Let f (x , y) = 4xy − x3 − 2y2

(b) Find the max value of f (x , y) on the unit disk {(x , y) | x2 + y2 ≤ 1}.

We know that the maximum value occurs on the boundary of the unitdisk.

How do we find where?

We know the boundary is parametrized by

x = cos(t), y = sin(t), z = f (cos(t), sin(t)).

=⇒ We want to maximize z = f (cos(t), sin(t)).




We know that the maximum value occurs on the unit disk, so we want tomaximize z = g(t) = f (cos(t), sin(t)).

Find where g ′(t) = 0!

First, need to find g ′(t).

g ′(t) =∂f

∂x

dx

dt+

∂f

∂y

dy

dt

=(4y − 3x2

)(− sin(t)

)+(4x − 4y

)(cos(t)

)=

(4 sin(t)− 3 cos2(t)

)(− sin(t)

)+(4 cos(t)− 4 sin(t)

)(cos(t)

)= −4 sin2(t) + 3 cos2(t) sin(t) + 4 cos2(t)− 4 cos(t) sin(t)

= 4 cos2(t)− 4 sin2(t) + 3 cos2(t) sin(t)− 4 sin(t) cos(t)


Solutions

2. Let f (x , y) = 4xy − x3 − 2y2



We know we want to find where g ′(t) = 0, and we have found that

g ′(t) = 4 cos2(t)− 4 sin2(t) + 3 cos2(t) sin(t)− 4 sin(t) cos(t).

Using Maple, I can solve for where g ′(t) = 0, and I find that there are 4possible values of t that could give us our absolute max on this region:

t ≈ −.92 t ≈ .68 t ≈ 2.06 t ≈ −2.71.





We have found that the only possible values of t for which the maximumof f on the boundary could occur are:

t ≈ −.92 t ≈ .68 t ≈ 2.06 t ≈ −2.71.

Plugging each of these in (again using Maple), we find that f is largest att ≈ −2.71.

But we wanted to know the x ’s and y ’s that gave us our max, so now Iplug this value of t into x = cos(t), y = sin(t), and I find that f attainsits absolute max on the unit disk at x ≈ −.91, y ≈ −.42.




We have found that maximum value of f on the unit disk occurs atapproximately (−.91,−.42).


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