MATH 12002 - CALCULUS I 2.7: Related Rates Part 1: Introduction - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §2.7: Related Rates Part 1: Introduction & Example Revisited

Professor Donald L. White

Department of Mathematical Sciences Kent State University

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Related Rates

In the previous lecture, we had the following example:

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? The question is asking for the rate of change of the length of the string at a certain point, given the rate of change of the horizontal distance between the flier and the kite. These rates of change are clearly related to each other. To solve the problem, we were able to first write the horizontal distance as a function of time, then find the relationship between the length of the string and horizontal distance in order to write the length of the string as a function of time. We then evaluated the derivative of this function at the appropriate time.

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Related Rates

More generally, we want to consider the following situation: Given the rate of change of one quantity (or more) at a certain point, find the rate of change of a related quantity at that point. In general, we may not be able to explicitly write the quantities themselves as functions of time as we did in the previous example. But if we can find the relationship between the quantities, we can differentiate implicitly with respect to time and get a relationship between their rates of change, which is all we need anyway. Let’s see how this works in a very slight variation of our previous example.

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Examples

Example 1

A kite is flying horizontally at an altitude of 30 feet. If the horizontal speed is 2 ft/sec when the length of the string is 50 feet, how fast must the string be let out at this point in order to maintain the altitude?

Solution

As we did before, we let x = x(t) = horizontal distance to the kite at time t, ℓ = ℓ(t) = length of the string at time t. Given: dx

dt = 2 ft/sec when ℓ = 50 feet.

Want: dℓ

dt when ℓ = 50 feet.

We need to relate x and ℓ and then take derivatives (implicitly) with respect to time in order to relate dx

dt and dℓ dt .

[Continued →]

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Examples

Example 1 Solution [continued]

We have the following situation at a given time:

✑✑✑✑✑✑ ✑ ✲ q q

FLIER

30 ft x ℓ

KITE

By the Pythagorean Theorem, 302 + x2 = ℓ2. Taking the derivative of each side with respect to t, we obtain 0 + 2x dx dt = 2ℓdℓ dt and so dℓ dt = x ℓ · dx dt . [Continued →]

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Examples

Example 1 Solution [continued]

We now have dℓ dt = x ℓ · dx dt , and we know that when ℓ = 50, dx

dt = 2.

Since 302 + x2 = ℓ2, when ℓ = 50, we have 302 + x2 = 502, and so x =

• 502 − 302 = 40.

Hence when ℓ = 50, dℓ dt = 40 50 · 2 = 8 5 = 1.6 ft/sec.

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