Volume of a torus The volume of a torus (doughnut or bagel) wi...

Question

Volume of a torus The volume of a torus (doughnut or bagel) with an inner radius of a and an outer radius of b is V=π²(b+a)(b−a)²/4.

b. Evaluate this derivative when a=6 and b=10.

Accepted Answer

Hello. In this video, we are going to be approaching the following related rates problem. So, we are told that the surface area of a cone is given by the formula S is equal to pi R multiplied by the quantity R plus the square root of R2 + H2, where R is the radius and H is the height of the cone. Calculate the value of the rate of change of the radius with respect to height, DRDH, if S is equal to 1400 pi, R is equal to 5, and H is equal to 12. So, our goal for this problem is to solve for the derivative DRDH. This derivative represents the rate of change of the radius with respect to height. So, let's go ahead and first take a look at the given function. The function is given to us as S is equal to pi R multiplied by the quantity R plus the square root of R2 plus H2. Now, before we take the derivative with respect to height, in order to solve for DRDH, let's go ahead and first simplify the given function. So, we are told that the surface area is equal to 1400 pi. We can go ahead and immediately plug that into our given function, allowing us to rewrite this as 1400 pi. equal to pi R multiplied by the quantity, R plus the square root of R2 + H2. Now, since we have a factor of pi on both sides of the equation, we can divide both sides of the equation by pi, allowing us to eliminate the pi term. Then what we can go ahead and do is we can go ahead and distribute the R into the quantity of R plus the square root of R2 plus H2. That will allow us to rewrite the equation as 1400 is equal to R2 plus R multiplied by the square root of R2 plus H2. So, now what we want to do is we want to try to get the derivative DRDH. In order to do this, first, what we need to do is we need to take the derivative of the equation with respect to H. So that is going to be our next step in the problem. We are going to use implicit differentiation to take the derivative with respect to H. Now, 1400 is a constant value on the left-hand side of the equation. The derivative of a constant is zero, so we are left with 0 on the left hand side. Now we will take the derivative of R2d with respect to H. The derivative of R squared will be 2R by the power rule, but since this is a derivative of R with respect to H, we will generate a DRDH derivative. Now, what about the derivative of R multiplied by the square root? Well, in order to take this derivative, we are going to have to use the product rule. So, we will take the original function R. Then multiply this to the derivative of the square root of R2 plus H2. In order to take this derivative, we will go ahead and use the chain rule. Since the square root Of R2 + H2 can be rewritten as R2 + H2 raised the power of 1/2, the outermost function will be the exponent in 1/2, so we will need to use the power rule. That will give us 1/2 multiplied by R2 + H2, raise the power of 1/2 minus 1, which will be negative 1/2. Then because of the chain rule, we will need to multiply this by the derivative of the innermost function. So, the derivative of R squared is 2 R, but again, since this is a derivative with respect to H, we generate a DR DH term. And the derivative of H2d will be 2 H. So this is going to be the first part of the product rule. Now, for the rest of the product rule, we will take the square root of R2 plus H2d. And multiply this to the derivative of R with respect to H. The derivative of R will be one, but again because of implicit differentiation, we will generate a DRDH term. So this is going to be the entirety of the given product rule. Now what we want to do is we want to go ahead and simplify the right-hand side of the equation. So If we take a look at the first terms of the product rule, I forgot to put a bracky here, apologies, we will go ahead and multiply all of the terms together. That is going to be R multiplied by the quantity to R DRDH plus 2 H divided by 2 multiplied by the square root of R2 plus H2. And on the right-hand side, we just have the square root of R squared plus H2d multiplied by DRDH. Notice that on this term of the product rule, every term in the numerator has a multiple of 2, and we have a multiple of 2 in the denominator. What this means is that we can factor out a 2. From the numerator and simplify that factor with 2 from the denominator. That will allow us to reduce the coefficients. To just be one. But now what we need to do is we need to multiply this R term into the quantity RDRDH plus H. When we multiply the R into the numerator, that is going to allow us to rewrite the numerator as R squared DRD H plus R H. This is going to be the simplified form of the product rule. And this is all still equal to 0. Now, what we want to do is we want to eliminate this denominator of the square root R2 plus H2. In order to do that, we can go ahead and multiply the entirety of the equation by the square root of R2 plus H2. When we do this, this will allow us to rewrite the equation to be 0, equal to 2R multiplied by the square root of R2 + H2, multiplied by DRDH. Plus R 2 DRDH plus RH And multiplied by R2 + H2 underneath the square root, multiplied by that quantity, that will allow us to eliminate the square root, leaving us with the quantity R2 plus H2 multiplied by DRDH. One thing that we can do to the last term is we can go ahead and distribute this DRDH term into R2 plus H2. That will allow us to rewrite the last term to be R2d multiplied by DRDH plus H2 multiplied by DRDH. Now, there are two terms that we can combine here. We can combine R2 DRDH with another R2 DRDH. That will allow us to rewrite the equation to be 0 equal to 2 R multiplied by the square root of R2 plus H2 multiplied by DRDH plus 2 R 2 DRDH plus R H plus H2 DRDH. OK, so this has been quite a bit to keep up with, but since this is the simplified form of the given derivative, notice that not every term in the derivative has a multiple of DRDH. Recall that our goal is to solve for DRDH. So what we are going to do is we are going to move the one term that doesn't have a DRDH multiple to the left hand side of the equation. So what we are going to do is we're going to subtract R H to both sides of the equation. That will leave us with negative RH equal to 2 R multiplied by the square root of R2 + H2 multiplied by DRDH plus 2 R 2 DRDH plus H2 DRDH. Now that every term on the right-hand side of this equation has a DRDH multiple, we can go ahead and factor out DRDH in order to solve for the derivative. That will leave us with DR DH multiplied by the quantity, 2 R multiplied by the square root of R2 plus H2 + 2 R 2 plus H2. Equal to negative RH. And finally, if we divide both sides of the equation by this quantity here, that will give us the derivative DRDH equal to negative RH divided by the quantity. 2 are multiplied by the square root of R2 plus H2 + 2 R 2 + H2, and this is going to be the final form of the given derivative. So, now that we've come up with an equation for our gold derivative, which is the rate of change in the radius with respect to height, what we want to do now is we want to solve for the value of the derivative at the specified quantities. Now, the problem told us earlier that we want to solve the value for this derivative when the radius is 5 and the height is 12. We will go ahead and plug in these values into the derivative. That is going to give us -5 multiplied by 12, divided by 2 multiplied by 5, multiplied by the square root of 5 squared. Plus 12 squared. Plus 2 multiplied by 5 squared. Plus 12 squared. And after simplifying this function, we will be left with the value negative 5 divided by 27, and this is going to be the value of the rate of change of the radius with respect to height when radius is equal to 5 and height is equal to 12. And with that being said, the solution to this problem is going to be D. So I hope this video helps you in understanding how to approach this problem, and I will go ahead and see you all in the next video.

Volume of a torus The volume of a torus (doughnut or bagel) with an inner radius of a and an outer radius of b is V=π²(b+a)(b−a)²/4.
b. Evaluate this derivative when a=6 and b=10.

Key Concepts

Volume of a Torus

Derivative

Evaluation of Functions

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