Written Homework 7 Solutions

Written Homework 7 Solutions
MATH 1310 - CSM
Assignment: pp 153-156 problem 6, 8, 10, 11, 13, 15, 17, 18, 20.
6. Find formulas for the derivatives of the following functions; that is, differentiate them.
Solution:
(a) f ′ (x) = 21x6 − 1.2x3 + 3πx2 (b) g ′ √
3
(x) =
2 √ 35
−
x x6 (c) h ′ (w) = 16w 7 − cos(w) − 2
3w3 (d) R ′ (u) = −4 sin(u) − 3 sec 2 (u) + 1
3u 2/3
(e) V ′ (s) = − 1
4s 3/4
(f) F ′ (z) = ln(2) √ 7 · 2 z
(g) P ′ (t) = −a · t + v0
8. In each case below, find a function f(x) whose derivative f ′ (x) is:
(a) f ′ (x) = 12x 11 ; Solution: f(x) = x 12
(b) f ′ (x) = 5x 7 ; Solution: f(x) = 5x8
8
(c) f ′ (x) = cos(x) + sin(x) ; Solution: f(x) = sin(x) − cos(x)
(d) f ′ (x) = ax 2 + bx + c ; Solution: f(x) = ax3
(e) f ′ (x) = 0 ; Solution: f(x) = 3
(f) f ′ (x) = 5
√ x ; Solution: f(x) = 10 √ x
3
+ bx2
2
10. (a) For which values of x is the function x − x 3 increasing?
Solution: If we want to know when x−x 3 is increasing, we should find its derivative,
set it equal to zero to find the critical points, and then test the intervals created by the
critical points to see when the derivative is positive. Remember that if the derivative
is positive, we know that the function is increasing.
+ cx
f(x) = x − x 3 , so f ′ (x) = 1 − 3x 2
1

Now set f ′ (x) = 0
1 − 3x 2 = 0
1 = 3x 2
1
3 = x2 , hence
1
x = ±
3
So we need to test the following intervals: (−∞, −
1
3
1 ), (− 3 ,
1
3 ) and (
1
3
, ∞).
By plugging in points from these intervals into f ′ (x), we can figure out the behavior
of f(x). These calculations have been omitted, but f ′ (x) > 0 for x ∈ (−
thus f(x) is increasing on this interval.
1
3 ,
1
3 ),
(b) Where is the graph of y = x − x3 rising most steeply?
Solution: We want to know when y = x − x3 is rising most steeply. We know it has
1
to be in the interval (− 3 ,
1
3 ) since this is the only time when f is increasing. One
way to figure out when f is rising most steeply is to look at the second derivative. If
f ′ (x) = 1 − 3x2 , then f ′′ = −6x. Now we set f ′′ (x) = 0 to find its critical points and
we see that f ′′ (x) = 0 when x = 0. Notice f ′ (0) = 1 > 0 so x = 0 is a maximum for
f ′ (x). So to answer our original question, f is rising most steeply when x = 0.
(c) At what points is the graph of y = x − x 3 horizontal?
Solution: We have already seen that the critical points of f are at x = ±
1
3
. Recall
that f obtains its critical points when f ′ (x) = 0. So f(x) is horizontal, or of zero
slope, or f ′ (x) = 0, when x = ±
1
(d) Make a sketch of the graph of y = x − x 3 that reflects all these results.
Solution:
3 .
-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5
2
1.6
1.2
0.8
0.4
-0.4
-0.8
-1.2
-1.6

11. (a) Sketch the graph of the function y = 2x + 5
on the interval 0.2 ≤ x ≤ 4.
x
Solution:
20
15
10
5
0 0.5 1 1.5 2 2.5 3 3.5
(b) Where is the lowest point on that graph? Give the value of the x-coordinate exactly.
Solution: The lowest point on this graph occurs when 0.2 ≤ x ≤ 4 and f ′ (x) = 0.
Since f(x) = 2x + 5
x , f ′ (x) = 2 − 5
, and after some simple algebra, we see that
x2 f ′
5
(x) = 0 when x = . Thus the lowest point on this graph is ( , 6.3246).
5
2
13. This problem was not graded and is very similar to 15.
15. (a) Write the microscope equation for y = √ x at x = 3600.
Solution: The microscope equation is y ≈ √ 3600 + △y = 60 − 1
· △x.
120
(b) Use the microscope equation to estimate √ 3628 and √ 3592. How far are these estimates
from the values given by a calculator?
Solution: Using the above microscope equation, we will first estimate √ 3628. To
figure out our estimate for y, we first need to find △x: △x = 3600 − 3628 = −28.
Now we can plug into y ≈ 60 − 1
1
· △x. So y ≈ 60 − · (−28) ≈ 60.23333. If we
120
plug √ 3628 into a calculator we get 60.23388, so there is very little discrepancy.
Now we can estimate √ 3592. We first need △x: △x = 3600 − 3592 = 8. Thus
y ≈ 60 − 1
1
· △x. So y ≈ 60 −
120 120 · (8) ≈ 59.93333. If we plug √ 3628 into a
calculator we get 59.933296, so again there is very little discrepancy.
17. A ball is held motionless and then dropped from the top of a 200 foot tall building. After
t seconds have passed, the distance from the ground to the ball is d = f(t) = −16t 2 + 200
feet.
3
120
2

(a) Find a formula for the velocity v = f ′ (t) of the ball after t seconds. Check that your
formula agrees with the given information that the initial velocity of the ball is 0
feet/second.
Solution: Since we’re given that v = f ′ (t) and f(t) = −16t 2 + 200, it’s easy to see
that v(t) = −32t. Now we can check that v0 = 0: v(0) = −32 · 0 = 0.
(b) Draw graphs of both the velocity and the distance as functions of time. What time
interval makes physical sense in this situation? (For example, does t < 0 make sense?
Does the distance formula make sense after the ball hits the ground?)
Solution: The time interval that makes the most sense is 0 ≤ t ≤ 4 since we think
of time starting at zero and the ball hits the ground shortly before t = 4.
160
80
0 0.4 0.8 1.2 1.6 2 2.4 2.8 3.2 3.6 4
-80
-160
y = -16x² + 200
y = -32x
(c) At what time does the ball hit the ground? What is its velocity then?
Solution: The ball hits the ground when d = f(t) = −16t 2 +200 = 0 which happens
when t ≈ 3.54. When t = 3.54, the ball’s velocity is v(3.54) = −32 · 3.54 = −113.14.
18. A second ball is tossed straight up from the top of the same building with a velocity of 10
feet per second. After t seconds have passed, the distance from the ground to the ball is
d = f(t) = −16t 2 + 10t + 200 feet.
(a) Find a formula for the velocity of the second ball. Does the formula agree with given
information that the initial velocity is +10 feet per second? Compare the velocity
formulas for the two balls; how are they similar, and how are they different?
Solution: Again, we know f(t) and that v = f ′ (t), so we get that v = −32t + 10.
(b) Draw graphs of both the velocity and the distance as functions of time. What time
interval makes physical sense in this situation?
Solution: Again, the time interval that makes the most sense is 0 ≤ t ≤ 4.
4

160
80
0 0.4 0.8 1.2 1.6 2 2.4 2.8 3.2 3.6 4
-80
-160
y = -16x² + 10x + 200
y = -32x + 10
(c) Use your graph to answer the following questions. During what period of time is the
ball rising? During what period of time is it falling? When does it reach the highest
point of its flight?
Solution: The ball is roughly rising for 0 < t < 0.31 and falling for 0.31 < t < 3.85.
The highest point is reached when v = −32t + 10 = 0 which occurs when t ≈ 0.31.
(d) How high does the ball rise?
Solution: By plugging t = 0.31 into our distance equation, we see that the ball
reaches f(0.31) = −16(0.31) 2 + 10(0.31) + 200 = 201.56 feet, or 1.56 feet above the
top of the building.
20. A steel ball is rolling along a 20-inch long straight track so that its distance from the
midpoint of the track (which is 10 inches from either end) is d = 3 sin(t) inches after t
seconds have passed. (Think of the track as aligned from left to right. Positive distances
mean the ball is to the right of the center; negative distances mean it is to the left.)
(a) Find a formula for the velocity of the ball after t seconds. What is happening when
the velocity is positive; when it is negative; when it equals zero? Write a sentence or
two describing the motion of the ball.
Solution: v(t) = d ′ (t) = 3 cos(t). When the velocity goes from positive to negative
and visa versa it is simply switching directions on the track from right to left and
left to right respectively. When the velocity equals zero, the ball is at its maximum
distance from the center. The ball is moving back and forth along the track reaching
its higher speeds towards the center of the track and pausing briefly when its distance
from the center is maximized.
(b) How far from the midpoint of the track does the ball get? How can you tell?
Solution: The maximum distance occurs when d = 3 sin(t) is maximized which
5

occurs when v(t) = 3 cos(t) = 0 or when t = π
. Thus the maximum distance is
2
3 sin( π
) = 3 · 1 = 3 inches.
2
(c) How fast is the ball going when it is at the midpoint of the track? Does it ever go
faster than this? How can you tell?
Solution: The first time the ball is in the center of the track after it has started
to move is when d = 3 sin(t) = 0 which occurs when t = π. The speed at t = π is
|v(π)| = |3 cos(π)| = |3 · (−1)| = | − 3| = 3, so the speed of the ball at the midpoint
of the track is 3 inches per second which is the maximum speed.
6