A sphere is growing at a rate of 50cm3s50\frac{\operatorname{\mathrm{cm}}^3}{s}50scm3. At what rate is the radius of the sphere increasing when the radius is 5cm5\operatorname{\mathrm{cm}}5cm?

0.159 c m ⁡ s 0.159\frac{\operatorname{\mathrm{cm}}}{s} 0.159 s cm

A sphere is growing at a rate of 50 c m 3 s 50\frac{\operatorname{\mathrm{cm}}^3}{s} 50 s cm 3 . At what rate is the radius of the sphere increasing when the radius is 5 c m 5\operatorname{\mathrm{cm}} 5 cm ?

A sphere is growing at a rate of 50 cubic centimeters per second. At what rate is the radius of the sphere increasing when the radius is 5 centimeters? So this is a related rates problem where we have a geometry or some sort of shape that is growing or shrinking. So that is what we're gonna be focusing on in this video. And so to do this, well, a good first step is gonna be to draw a picture of the situation. So you can imagine we have this three-dimensional sphere, and this sphere has a radius of at one instance of 5 centimeters. Now I wanna be clear here. This is not always gonna be the radius of the sphere. Just the instant that we're looking at, that's going to be a situation where the radius is 5 centimeters. This whole thing is growing outward, and we wanna know at what rate is the radius of the sphere changing. Well, now that we have a picture of the situation, what I need to think about is what equation is going to work to solve this problem. And since I can see that we're dealing with a sphere and we're given the radius and we're told that it's increasing, I think it makes sense that the volume of a sphere would be the equation that you need. Now the volume of the sphere is 4 thirds pi times the radius of the sphere cubed. Now this equation alone is not gonna allow us to just solve this problem instantly because we have rates in this problem. The sphere is growing at a rate of of 50 cubic centimeters per second, and we're looking for the rate at which the radius of the sphere is increasing. Now what I can see is that we have the radius at one instant, which is 5 centimeters, and we also have the rate at which the volume of this entire thing is growing. So I'm gonna say that the volume over time, d v over d t, is growing at 50 cubic centimeters per second. But what we're looking for is the rate at which the radius is growing and that's going to be d r over d t, the changing radius over time, which is what we don't have. So let's go ahead and find this by first taking the derivative on both sides of our volume equation. And we know to do this, we're in essence just taking this time derivative of every single variable that we have. So taking the time derivative of volume, we get d v over d t. Taking the derivative of this portion right here, well, we can use the power rule. So we can multiply this 3 by the front, reduce the power by 1. These are all just constants, so we don't have to worry about them, which means we're going to get 4 thirds pi. And then using the power rule, we're going to get 3 times the radius squared because that's just bringing this power to the front and reducing it by 1. But then we need to multiply this by the rate d r over dt using the chain rule since we just took the derivative of r with respect to time. Now notice in this situation, we can cancel this 3 with that one since that's in in a denominator, that's in a numerator, or just being multiplied here. So what that means is that our rate d v over d t is going to be equal to 4 pi times the radius squared multiplied by d r over d t. Now at this point, what I'm trying to do is solve for the rate at which the radius of the sphere is increasing. That's the d r over d t that we're trying to find. So that means that I need to isolate this on one side of the equal sign. So I'm going to divide 4 pi r squared on both sides of this equation to get the rate for the radius by itself. So that's going to cancel all these terms right here. Now what that's going to give me is that d r over d t, which I'll write over here, is going to be equal to d v over d t, this rate on top, divided by 4 pi r squared. And that's going to be our equation. Now from here, I can just plug in the variables that we know. So we know that dv over dt is 50, and then that's going to be divided by 4 times pi times the radius squared, which in this case, the radius is 5 centimeters. That's gonna be 5 squared. And then what you can do is you can either put this onto a calculator or you can do this by hand. If you do this by hand, you should get that the whole thing comes out to 1 over 2 pi. And on a calculator, that's approximately equal to 0.159. So that's what our result comes out to. And since we have a radius that is increasing over time, well, a radius is gonna be measured in centimeters and the time is gonna be measured in seconds. So that is going to be our rate d r over d t, and that's the solution to this problem. So either one of these answers that you see, 1 over 2 pi or this decimal answer, are both going to give you the correct solution. So that is going to be how you can solve this type of problem. Hope you found this video helpful.

A sphere is growing at a rate of 50 c m ⁡ 3 s 50\frac{\operatorname{\mathrm{cm}}^3}{s} 50 s cm 3 . At what rate is the radius of the sphere increasing when the radius is 5 c m ⁡ 5\operatorname{\mathrm{cm}} 5 cm ?

0.159 c m ⁡ s 0.159\frac{\operatorname{\mathrm{cm}}}{s} 0.159 s cm

1.57 c m s ⁡ 1.57\operatorname{\frac{\mathrm{cm}}{s}} 1.57 s cm

0.637 c m s ⁡ 0.637\operatorname{\frac{\mathrm{cm}}{s}} 0.637 s cm

0.318 c m s ⁡ 0.318\operatorname{\frac{cm}{s}} 0.318 s cm

4. Applications of Derivatives

Related Rates

Calculus

4. Applications of Derivatives

Related Rates

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Multiple Choice

A sphere is growing at a rate of $50\frac{\operatorname{\mathrm{cm}}^3}{s}$ . At what rate is the radius of the sphere increasing when the radius is $5\operatorname{\mathrm{cm}}$ ?

$1.57\operatorname{\frac{\mathrm{cm}}{s}}$

$0.637\operatorname{\frac{\mathrm{cm}}{s}}$

$0.159\frac{\operatorname{\mathrm{cm}}}{s}$

$0.318\operatorname{\frac{cm}{s}}$

Verified step by step guidance

Start by recalling the formula for the volume of a sphere: $ V = \frac{4}{3} \pi r^3 $, where $ V $ is the volume and $ r $ is the radius.

Differentiate the volume formula with respect to time $ t $ to find the relationship between the rate of change of volume $ \frac{dV}{dt} $ and the rate of change of radius $ \frac{dr}{dt} $. This gives: $ \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} $.

Substitute the given rate of change of volume $ \frac{dV}{dt} = 50 \frac{\mathrm{cm}^3}{s} $ and the radius $ r = 5 \mathrm{cm} $ into the differentiated equation.

Solve for $ \frac{dr}{dt} $ by rearranging the equation: $ \frac{dr}{dt} = \frac{\frac{dV}{dt}}{4 \pi r^2} $.

Calculate $ \frac{dr}{dt} $ using the substituted values to find the rate at which the radius is increasing.

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