MATH 12002 - CALCULUS I 2.3 (Part 3): Rates of Change - More - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §2.3 (Part 3): Rates of Change - More Examples

Professor Donald L. White

Department of Mathematical Sciences Kent State University

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Examples

Example

Suppose I throw a watermelon straight up from a tower that is 96 feet high at a velocity of 16 feet per second. The distance of the watermelon above the ground t seconds later is s(t) = −16t2 + 16t + 96 feet. How long does it take for the watermelon to reach the ground and at what velocity does it strike the ground?

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Examples

Solution

When the melon hits the ground, the distance above the ground is 0 feet, and so it reaches the ground at the time t when s(t) = 0. Therefore, 0 = −16t2 + 16t + 96 = −16(t2 − t − 6) = −16(t + 2)(t − 3), and so t = 3 or t = −2. But t cannot be negative, so the melon hits the ground after t = 3 seconds. We know that the velocity at time t is given by v(t) = s′(t) = −32t + 16 feet per second. Hence the velocity of the melon when it hits the ground is the velocity at time t = 3, which is v(3) = −32(3) + 16 = −96 + 16 = −80 feet per second.

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Examples

Example

A kite is flying horizontally at a constant speed of 2 feet per second at an altitude of 30 feet. How fast must the string be let out when the length of the string is 50 feet in order to maintain the altitude?

Solution

Think of the kite starting directly over the flier, and let x be the horizontal distance from the flier t seconds later. Also, let’s set ℓ equal to the length

• f the string after t seconds.

Both ℓ = ℓ(t) and x = x(t) are functions of time t, and we want to find the rate of change of the length ℓ with respect to time t when ℓ = 50, i.e., dℓ

dt when ℓ = 50 feet.

Since the horizontal speed is 2 feet per second, we have that x(t) = 2t. It is clear that x and ℓ are related, but how? If we can write ℓ as a function

• f x, then we can write ℓ as a function of t and find dℓ

dt .

[Continued →]

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Examples

Solution [continued]

We have the following situation at a given time:

✑✑✑✑✑✑ ✑ ✲ q q

FLIER

30 ft x ℓ

KITE

By the Pythagorean Theorem, x2 + 302 = ℓ2, hence ℓ(t) =

• x(t)2 + 900.

As we saw above, x(t) = 2t, so ℓ(t) =

• (2t)2 + 900 =
• 4t2 + 900.

Therefore, ℓ′(t) = 1

2(4t2 + 900)−1/2 · 8t =

4t √ 4t2 + 900 . [Continued →]

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Examples

Solution [continued]

We need ℓ′(t) at the time t when ℓ = 50 feet. Note that when ℓ = 50 feet, x2 + 302 = 502 and so 2t = x =

• 502 − 302 = 40.

Hence when ℓ = 50, we have t = 20, and the string must be let out at a speed of ℓ′(20) = 4 · 20

• 4(20)2 + 900

= 80 √ 2500 = 80 50 = 1.6 feet per second.

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