MATH 12002 - CALCULUS I 2.7: Related Rates Part 2: Examples - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §2.7: Related Rates Part 2: Examples

Professor Donald L. White

Department of Mathematical Sciences Kent State University

D.L. White (Kent State University) 1 / 6

Examples

Example 2

Sand is being poured onto a pile at a rate of 1.5 cubic feet per minute in such a way that it forms a cone whose height is equal to its radius. How fast is the height increasing when the pile is 5 feet high?

Solution

We are given the rate of change of volume of the sand and need to determine the rate of change of the height of the pile. We let V = volume of the sand at time t, h = height of the pile at time t. Given: dV

dt = 1.5 ft3/min.

Want: dh

dt when h = 5 ft.

We need to relate V and h and then take derivatives with respect to time in order to relate dV

dt and dh dt .

[Continued →]

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Examples

Example 2 Solution [continued]

The volume of a cone is given by V = 1

3πr2h, where r is the radius of the

(circular) base and h is the height. We are given that the height of the pile is equal to its radius, hence r = h and V = 1 3πh3. Taking derivatives with respect to time, we have dV dt = 3 · 1 3πh2 dh dt = πh2 dh dt . Therefore, when h = 5, dh dt = 1.5 π · 52 = 3 50π ft/min ≈ .019 ft/min.

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Examples

Example 3

A helium balloon is released on the ground 35 feet from an observer and rises vertically. If the angle between the line of sight and the ground is increasing at a rate of 0.5 radians per minute, how fast is the balloon rising when it is 49 feet off the ground?

Solution

We let θ = angle between line of sight and ground at time t, B = altitude of the balloon at time t. Given: dθ

dt = 0.5 rad/min.

Want: dB

dt when B = 49 ft.

We need to relate B and θ and then take derivatives with respect to time in order to relate dB

dt and dθ dt .

[Continued →]

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Examples

Example 3 Solution [continued]

We have the following situation at a given time:

✑✑✑✑✑✑ ✑ ✻ q q

OBSERVER θ 35 ft B BALLOON

Since this is a right triangle, we have tan θ = B 35, or B = 35 tan θ. Taking derivatives with respect to time, we obtain dB dt = 35(sec2 θ)dθ dt . [Continued →]

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Examples

Example 3 Solution [continued]

We have dB dt = 35(sec2 θ)dθ dt and dθ

dt = 0.5, so we need to determine sec2 θ when B = 49.

Recalling that sec = hyp

adj and using the Pythagorean Theorem, we have

sec θ = √ 352 + 492 35 = √ 72 · 52 + 72 · 72 35 = 7 √ 52 + 72 35 = √ 74 5 . Therefore, when B = 49, dB dt = 35 √ 74 5 2 (0.5) = 35 · 74 25 · 2 = 7 · 37 5 = 259 5 = 51.8 ft/min.

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