The Fundamental Theorem of Calculus Exercises

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If negative velocity means the cheetah is running South, then positive velocity means the cheetah is running North.

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The absolute value of -60 is 60, so Jen is going 60 mph.

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If the snail's velocity is 0 feet per second, the snail's speed is also 0 feet per second. This means the snail isn't moving at all.

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The bug is moving left at a speed of 4 units per second.

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(a) The hummingbird is moving neither North nor South whenever v(t) = 0, which occurs at t = 2, 5, and 7.

(b) The hummingbird is moving North when v(t) is positive, which occurs on the interval (2, 5).

(c) The hummingbird is moving South when v(t) is negative, which occurs on the intervals [0, 2) and (5, 7). We do include t = 0 in the first interval, since v(0) is negative.

(d) The hummingbird is moving most rapidly North when v(t) is positive and as large as possible. This is when t = 4.

(e) The hummingbird is moving most rapidly South when v(t) is at its most negative. This occurs at t = 6.

(f) The hummingbird is moving most rapidly when |v(t)| is largest. This occurs when t = 4 or 6.

(g) The hummingbird is speeding up on all intervals where v(t) is getting farther from zero (this means its speed |v(t)| is getting bigger). For this graph, v(t) is getting farther from zero on the intervals (2, 4) and (5, 6).

(h) The hummingbird is slowing down on all intervals where its speed |v(t)| is getting smaller, meaning v(t) is getting closer to zero. In this case, v(t) is getting closer to zero on the intervals (0, 2), (4, 5), and (6, 7).

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(a) The bug changes direction whenever v(t) goes from positive to negative, which happens at t = 4 and t = 12, or from negative to positive, which happens at t = 8. We do NOT include t = 0, because we can't tell if v(t) is negative to the left of t = 0. To summarize, the bug changes direction at t = 4, 8, and 12.

(b) The bug is moving to the right whenever v(t) is positive, which occurs on the intervals (0,4) and (8,12).

(c) The bug is moving to the left whenever v(t) is negative, which occurs on the intervals (4,8) and (12,16).

(d) The bug is moving most rapidly to the right when t = 10, since this is where the largest positive value of v(t) occurs.

(e) The bug is moving most rapidly to the left when t = 16, since this is where the most negative value of v(t) occurs.

(f) The bug is moving most rapidly when t = 16, since this is where |v(t)| is largest (equivalently, this is where v(t) is farthest from 0).

(g) The bug is speeding up on (0,2), (4,6), (8,10), and (12,16), since these are the intervals where v(t) is getting further from 0.

(h) The bug is slowing down on (2,4), (6,8), and (10,12), since these are the intervals where v(t) is getting closer to 0.

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(a) The cat changes direction whenever its velocity changes from positive to negative or from negative to positive. The only time this happens on [0,7] is at t = 5, where the cat changes from climbing up the tree (positive velocity) to climbing down the tree (negative velocity).

(b) The cat is climbing most rapidly upwards when t = 0 and when t = 4, since these are the times when the cat's velocity is largest.

(c) The cat is climbing most rapidly downwards when t = 6, since this is when the cat's velocity is most negative.

(d) The cat's fastest speed on the interval [0,7] is 2 feet per second, which occurs at t = 0 and t = 4.

(e) The slowest the cat ever goes on [0,7] is 0 feet per second.

(f) From t = 0 to t = 5 the cat is climbing up, and from t = 5 to t = 7 the cat is climbing down. The cat is highest in the tree when t = 5, since this is when it stops climbing up and starts climbing down.

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From t = 0 to t = 3 the bug travels

to the right.

From t = 3 to t = 5 the bug travels a weighted distance of

,

meaning it travels 3 units to the left.

From t = 5 to t = 8 the bug travels

to the right.

(a) At t = 0 the bug is at 0 on the numberline.

From t = 0 to t = 3 the bug moves 5 units right, so at t = 3 the bug is at 5.

From t = 3 to t = 5 the bug moves 3 units left, so at t = 5 the bug is at 2.

From t = 5 to t = 8 the bug moves 10 units right, so the bug ends up at 12 on the number line when t = 8.

(b) The bug will be moving the same distances, but with a different starting place.

At time t = 0 the bug is at -10. From t = 0 to t = 3 the bug moves right 5, so when t = 3 the bug is at

-10 + 5 = -5.

From t = 3 to t = 5 the bug moves left 3, so when t = 5 the bug is at

-5 – 3 = -8.

From t = 5 to t = 8 the bug moves right 10, so when t = 8 the bug is at

-8 + 10 = 2.

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(a) From time t = 0 to t = 4 the car is traveling east because the graph of v(t) is positive on (0, 4).

(b) From time t = 6 hours to time t = 10 hours the car is traveling west, because the graph of v(t) is negative on (6, 10).

(c) The car is stopped whenever its velocity is 0, which occurs at t = 4, 6, and 10 hours.

(d) By looking at the graph of v(t), we can see that from t = 0 to t = 6 the car traveled a total of 120 miles east:

From t = 6 to t = 10 the car travelled 80 miles west:

The car's net change in position was

120 – 80 = 40

miles to the east, so the car ended up 40 miles east from where it started.

Answer

First we need to understand the set-up. A toddler is between a lollipop and a teddy bear, each of which is 10 feet away from the toddler. We put the teddy bear on the right side because the problem says positive velocities mean the toddler is moving towards the teddy bear. We're used to thinking of positive velocities as moving us to the right on a number line, so if we put the teddy bear on the right side of the room, positive velocity still corresponds to moving right.

(a) We look at the graph to see how far the toddler moves over each time interval.

From t = 0 to t = 2 seconds the toddler moves

to the right, so the toddler is 14 feet from the lollipop and 6 feet from the teddy bear:

From t = 2 to t = 4 seconds the toddler moves

which means it moves 2 feet to the left (closer to the lollipop). The toddler is now 12 feet from the lollipop and 8 feet from the teddy bear.

From t = 4 to t = 6 seconds the toddler moves

closer to the bear, so the toddler is now 15 feet from the lollipop and 5 feet from the bear.

From t = 6 to t = 10 the toddler moves

meaning 6 feet to the left (closer to the lollipop). The toddler is now 9 feet from the lollipop and 11 feet from the bear.

What we did here was put the lollipop, toddler, and bear on a number line. The lollipop was at -10, the toddler at 0, and the bear at 10.

There are other ways to do this problem. For example,

we could put the lollipop at 0, the bear at 20, and start the toddler off at 10.

Then the toddler would end up at

10 + 4 – 2 + 3 – 6 = 9

on the numberline, meaning 9 feet from the lollipop and 11 feet from the bear.

Another possibility is that we could put the lollipop at -20, the bear at 0, and the toddler at -10:

Any way we do it, the toddler will end up 9 feet from the lollipop and 11 feet from the bear.

(b) The toddler is at the center of the room when t = 0, and passes through the center of the room once more between t = 6 and t = 10, in order to get closer to the lollipop than to the bear. This means the toddler passes through the center of the room twice.

(c) The toddler never reaches the lollipop or the bear. The toddler gets within 5 feet of the bear, and within 9 feet of the lollipop, but never gets to either.