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Practice Exam # 2 (2.73.9) (6968235)
Due: Thu Mar 26 2015 11:59 PM PDT
1. Question Details SCalcET7 2.8.027. [1735792]

+
lim h → 0
lim h → 0
lim h → 0
−1 2 1 − x
The graph of f is given. State the numbers at which f is not differentiable. x= 4 (smaller value) x= 0 (larger value)
Solution or Explanation
Consider the following function.
if |x| > 3
if |x| < 3
For what values of x is the function not differentiable? (Enter your answers as a commaseparated list.)
x =

f'(x) =
3/21/2015 Assignment Previewer
(a) Note that for
To show that does not exist we investigate by computing the left and righthand derivatives.
Since the left and right limits are different, does not exist, that is, does not exist. Similarly,
does not exist. Therefore, f is not differentiable at 3 or at −3.
(b)

x2 − 9 < 0 x2 < 9 ⇔ |x| < 3 ⇔ −3 < x < 3. So
f(x) = f'(x) = = x2 − 9 if x ≤ −3
−x2 + 9 if −3 < x < 3
x2 − 9 if x ≥ 3
2x if x < −3 −2x if −3 < x < 3 2x if x > 3
2x if |x| > 3 −2x if |x| < 3
f'(3) lim h → 0
f(3 + h) − f(3) h
f'+(3) = = = = (6 + h) = 6.
lim h → 0−
h → 0− [−(3 + h)2 + 9] − 0
h lim h → 0−
h → 0+ [(3 + h)2 − 9] − 0
h lim h → 0+
Find the derivative of the function below in two ways.
(a) by using the Quotient Rule
F'(x) =
(c) Which method appears to be simpler for this problem?
Differentiate. (Assume c is a constant.)
z' =
Differentiate.
Find an equation of the tangent line to the curve at the given point.
y =
Find the derivative of the function.
y' =
y = y' = (5 + 8e9x)−1/2 (5 + 8e9x) = (8e9x · 9) = 5 + 8e9x 1 2
d dx
y' =
y' =
f '(t) =
y' =
Find dy/dx by implicit differentiation.

Find dy/dx by implicit differentiation.
Differentiate the function.
f'(x) =
f(x) = log10(x9 + 7) f'(x) = (x9 + 7) = 1 (x9 + 7)ln 10
d dx
Use logarithmic differentiation to find the derivative of the function.
y' =
y = x − 4 x8 + 5 ln y = ln x − 4
x8 + 5
2 1 2
y' = − · 8x71 y
4x7
Use logarithmic differentiation to find the derivative of the function.
y' =
The height (in meters) of a projectile shot vertically upward from a point 4 m above ground level with an initial velocity of 21.5 m/s is after t seconds. (Round your answers to two decimal places.)

(b) When does the projectile reach its maximum height? 2.19 s
(c) What is the maximum height? 27.58 m
(d) When does it hit the ground? 4.57 s
(e) With what velocity does it hit the ground? 23.25 m/s
(a) the velocity after 2 s is and after 4 s is
(b) The projectile reaches its maximum height when the velocity is zero. ⇔ ⇔
(c) The maximum height occurs when
(d) The projectile hits the ground when ⇔ ⇔

v(2) = 1.9 m/s v(4) = 17.7 m/s
h(t) = 4 + 21.5t − 4.9t2 v(t) = h'(t) = 21.5 − 9.8t. v(2) = 21.5 − 9.8(2) = 1.9 m/s v(4) = 21.5 − 9.8(4) = −17.7 m/s.
v(t) = 0 21.5 − 9.8t = 0
t = = 2.19 s.21.5 9.8
t = 2.19. h(2.19) = 4 + 21.5(2.19) − 4.9(2.19)2 = 27.58 m.
h = 0 4 + 21.5t − 4.9t2 = 0 t = −21.5 ± 2(−4.9) 21.52 − 4(−4.9)(4)
t = tf ≈ 4.57 s [since t ≥ 0].
t = tf. v(tf) = 21.5 − 9.8tf ≈ −23.25 m/s [downward].
21. Question Details SCalcET7 3.9.005.MI. [1646152]
A cylindrical tank with radius 7 m is being filled with water at a rate of 4 m3/min. How fast is the height of the water increasing?
m/min
At noon, ship A is 170 km west of ship B. Ship A is sailing east at 40 km/h and ship B is sailing north at 15 km/h. How fast is the distance between the ships changing at 4:00 PM?
km/h
Given: at noon, ship A is 170 km west of ship B ship A is sailing east at 40 km/hr and ship B is sailing north at 15 km/h. If we let t be time (in hours), x be the distance traveled by ship A (in km), and y be the distance traveled by ship B (in km), then we are given that and

dz/dt t = 4h.
z2 = (170 − x)2 + y2 2z = 2(170 − x) − + 2y dz dt
dx dt
dy dt
x = 4(40) = 160 and y = 4(15) = 60 z = = 10 .(170 − 160)2 + 602 37
= (x − 170) + y = = dz dt
1 z
dx dt
dy dt
A man starts walking north at 2 ft/s from a point P. Five minutes later a woman starts walking south at 3 ft/s from a point 500 ft due east of P. At what rate are the people moving apart 15 min after the woman starts walking? (Round your answer to two decimal places.)
4.98 ft/s
Name (AID): Practice Exam # 2 (2.73.9) (6968235) Submissions Allowed: 100 Category: Homework Code: Locked: Yes Author: Mkrtchyan, Tigran ( [email protected] ) Last Saved: Mar 21, 2015 03:01 AM PDT Permission: Protected Randomization: Person Which graded: Last
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A kite 100 ft above the ground moves horizontally at a speed of 4 ft/s. At what rate is the angle (in radians) between the string and the horizontal decreasing when 200 ft of string have been let out?
1/100 rad/s

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