The volume V of a sphere of radius r changes over time t.

Question

c. At what rate is the radius changing if the volume increases at 10 in³ when the radius is 5 inches?

Accepted Answer

Welcome back, everyone. The volume V of a balloon is expanding as it is being inflated. If the volume increases at a rate of 15 centimeters cubed per second, when the radius R is 6 centimeters, at what rate is The radius increasing. We're given 4 answer choices A says 1 divided by 12, B 1 divided by 65, C 5 divided by 48, and D 1 divided by 4 pi all given in centimeters per second. So for this balloon, we're going to consider it as a sphere. So let's recall that the volume of a sphere V is 4/3 by our cubed where R is radius. What we're going to do is understand that our volume is changing, so it is a function of time because relative to time, this volume is increasing since the balloon is expanding. And we also know that radius R is changing, so it is a function of time because relative to time our radius must also be increasing, right? So what we're going to do is consider the given function and differentiate it with respect to time. Taking the derivative of the left side, we're going to get DV divided by DT, the derivative of volume. So essentially we're showing that DV divided by DT is the rate of change of volume with respect to time. Now on the right hand side we have 4/3 pi, which is a constant, and now R is cubed. So this means that first of all, we're applying the power rule which gives us 3 R squared. We need to bring down the exponent and decrease the original exponent by 1 unit, right? This is what the power rule says. And now applying the chain rules. We are going to get 3 R squared multiplied by DR divided by DT, so the rate of change of radius. What are we looking for? Well, we want to solve for DR divided by DT, but before we identify this term, we can simplify the expression. We're going to say that DV divided by DT is simply equal to. 4 pi R squared DR divided by DT and now solving 4 DR divided by DT. We can simply divide both sides by or by R squared. So we're going to get 1 divided by or R2, multiplied by DV divided by DT. This would be the rearranged expression for DR divided by DT and all we have to do is substitute the given data. So DR divided by DT is going to be equal to. And now we have to begin with the first term. So we have one in the numerator. In the denominator there's 4 pi, which is multiplied by r quad. The radius is given as 6 centimeters at the point of interest, so we square 6 centimeters, and then we multiply that by the weight of change of volume, which is 15 centimeters cubed per second. Now let's simplify. We're going to get 1 divided by 4 pi multiplied by 36, which is 144 centimeters squared at the bottom, and now at the top we have 15, right? centimeters cubed per second. Now we need to simplify this expression and in order to. Simplify it, we have to notice that the numerator and the denominator, those are both divisible by 3. So dividing the top by 3, we're going to get 5. Dividing the bottom by 3, we're going to get 48 i. What about the units? Well, we get centimeters per second. So the final answer to this problem corresponds to the answer choice C. Thank you for watching.

The volume V of a sphere of radius r changes over time t.
c. At what rate is the radius changing if the volume increases at 10 in³ when the radius is 5 inches?

Key Concepts

Related Rates

Volume of a Sphere

Differentiation

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