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A street light mounted at the top of a $ 15-ft-ta…

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Problem 14 Easy Difficulty

If a snowball melts so that its surface area decreases at a rate of $ 1cm^2/min, $ find the rate at which the diameter decreases when the diameter is $ 10 cm. $
(a) What quantities are given in the problem?
(b) What is the unknown?
(c) Draw a picture of the situation for any time $ t. $
(d) Write an equation that relates the quantities.
(e) Finish solving the problem.


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Related Courses

Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 3

Differentiation Rules

Section 9

Related Rates

Related Topics

Derivatives

Differentiation

Discussion

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ML

Morgan L.

June 27, 2022

L is the diameter. L = 2r

sk

Savan K.

April 23, 2022

I didn't understand anything from what he is saying what is L and from were he got it

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In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

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Video Transcript

The problem is if a snowball melts so that its surface area decreases at a rate of 100 centimeter square per meter, finding the rate at which the diameter decreases when determent is 10 centimeters and what quantities are given in the problem. Decreasing rate of the surface area and the length of the diameter what is unknown, but the cells decreasing rate of demeter see draw a picture of the situation for in time to for a ball. This is a mis a tent time we have. This is a picture of the situation for in time, t is right and equation that relates the quantities but surface area s is equal to pi times. L is a diameter, so this is equal to pi l square e, so differentiate this equation. Each side, respect to t behalf e s. T is equal to 2 pi l, l d t so behalf e o d is equal to 1 over 2 pi l times d s d, so this is equal to 1 over 25 times at this is 1 over 14 times. Pi.

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Calculus: Early Transcendentals

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