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A stone is dropped into a lake, creating a circular ripple that travels outward at a speed of 60 cm/s. Find the rate at which the area within the circle is increasing after (a) 1 s, (b) 3 s, and (c) 5 s. What can you conclude?
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01:28
Amrita Bhasin
Calculus 1 / AB
Chapter 3
Differentiation Rules
Section 7
Rates of Change in the Natural and Social Sciences
Derivatives
Differentiation
Missouri State University
Campbell University
Harvey Mudd College
University of Michigan - Ann Arbor
Lectures
04:40
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.
44:57
In mathematics, a differentiation rule is a rule for computing the derivative of a function in one variable. Many differentiation rules can be expressed as a product rule.
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A stone is dropped into a …
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A stone is dropped in a la…
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Area of a Ripple A stone i…
$$\begin{array}{l}{\te…
in this problem. We have a circular ripple in a lake, and we're interested in finding the rate of change of different times. And when they say the rate, they want the change in area with respect to time. We know the area as a function of radius, not as a function of time. But we also know that rate at which the radius is increasing because it says that the ripple travels outward at the speed of 60 centimeters per second. So what we can do is make this a composite function where the area as a function of time, would be pi times 60 t squared we can insert are 60 t and for our and let's go ahead and simplify that. And we have a of T equals 3600 pie t squared. Now we confined the derivative of that, and that will be the rate we're looking for. So a prime of tea is 7200 pie t. All right, we're going to find for part a the rate for one second and for part B, the rate for three seconds and for part c, the rate for five seconds. So if a prime of one, a prime of three in a prime of five. So we're going to substitute those values of t into are a prime function. So we have 7200 pi times one which is 7200 pie and the units would be square centimetres per second. And for 3 72 100 pi times three gives us 21,600 pie centimetres per second and then for 5 72 100 pi times five gives us 36,000 pie centimetres per second. So I was conclude that this area is growing large very quickly.
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