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A rubber ball (a sphere) is expanding in such a way that its radius increases from $8 \mathrm{cm}$ to $8.04 \mathrm{cm}$(a) What is the approximate change in its volume?(b) Is the approximate change larger or smaller than the actual change?$$\left(V=\frac{4}{3} \pi r^{3}\right).$$

(a) $10.24 \pi \approx 32.1699 \mathrm{cm}^{3}$(b) small

Calculus 1 / AB

Chapter 3

Applications of the Derivative

Section 6

Linearization and Differentials

Derivatives

Missouri State University

Harvey Mudd College

University of Nottingham

Idaho State University

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.

30:01

In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (the rate of change of the value of the function). If the derivative of a function at a chosen input value equals a constant value, the function is said to be a constant function. In this case the derivative itself is the constant of the function, and is called the constant of integration.

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in this question we are given a rubber ball. It has increased in volume from A radius of eight centimetres to these radios Of 8.04. Now the volume is given by for over three pi r cubed. And we have to calculate the change in volume which is total fee. Now first we differentiate T. V. To uh Which by the parlor the tree comes down. We have four pi R squared is D V. D. R. And now to calculate the change in V. You know that that will be four pi R squared. Apply by the change in our geo is coming up. He is substituting our values. We have T V Seiko to TV. Is it data? V is equal to four pie. Oh dear old radius which is um eight squared more played by the changing area. Now the change in the era is given by the old the old radius which is 8cm uh minus the new radios which is uh 0.8.04. Which We can also just know that it's an increase. So we subtract 8.04 eight is he go to? Now this one was sold to four pi Applied by 64, multiplied by 0.04. And This data is he will actually compute to 10 two, 10.24 pie cubic centimetres. Now this is the change in you know, you know radius which we can also have been In a few numbers and 32.16 99 cubic centimetres. Now the next question is to is this uh number, is this approximated? Changing volume? Actually smaller or bigger than the real or actual um Changing volume. Now to find the actual changing volume, we have to calculate the initial volume first. The initial the initial volume was for over three pie. We'll apply by the initial um radius is eight square eight cubed. It's cute. Now here we can find the initial volume as everyone is equal to 2176.99 sq cm. And r. V two is you go to 4/3 pie Are acute which is 8.04 cubed. And this computer to 2.1 four four six point 2.12 2144. 2008. 2144 .66. Actually this is the first folio. This is the okay, let me exchange there so we have our view one physical too. Mhm. Yeah. Yeah. You didn't make sense. 244 66 Here. And for the V two is 2176. Since it's expanding cubic centimeters. Now to find the actual, they change the actual changing volume. He say V two see one. and uh we get our actual change in volume is 32 .329 cubic centimeters. Now clearly we can see that this is bigger than the approximated. Um So answering the question, the approximated value is smaller. Mhm. Mhm

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