Surface area of a cone The lateral surface area of a cone of r...

Question

Surface area of a cone The lateral surface area of a cone of radius r and height h (the surface area excluding the base) is A = πr√r²+h².

a. Find dr/dh for a cone with a lateral surface area of A=1500π.

Accepted Answer

Hello. In this video, we are going to be approaching the total surface area problem. So we are told that the total surface area of a cylinder having radius R and a high H is given as A is equal to 2 piR multiplied by R plus H. We want to calculate the value of DRDH if A is equal to 500 pi. So, let's go ahead and break down the pieces of information that is given to us. First, we are given the equation of the total surface area of a of a cylinder. That area is A is equal to 2 piR multiplied by the quantity R plus H. We want to determine the derivative DRDH when the area is equal to 500 pi. So we are asked to find the derivative with respect to H, given the area 500 pi. Let's go ahead and first plug in this area, 500 pi. That will replace the given equation with 500 pi equal to 2 pi R multiplied by R plus H. To further simplify this equation, we can distribute 2R into the quantity R plus H, and that will allow us to rewrite this equation as 500 pi equal to 2 R 2 plus 2 R H. Now what we want to do is we want to go ahead and calculate the derivative of this equation with respect to H. So, on the left-hand side, we have 500 i. Pi is a constant value, and multiplying that value with 500 will produce another constant value. The derivative of a constant value is going to be zero, so the derivative of the left-hand side of this equation will just be 0. Now what we want to do is on the right hand side, we want to take the derivative with respect to H. So, The first term that we will take the derivative with respect to H is 2 pi R squared. We will take the derivative of R by using the power rule that will give us 2 pi multiplied by the derivative of Rsquared, which is 2R. But since we took the derivative of a function of R with respect to H by implicit differentiation, that is going to generate a DRDH term. We will go ahead and take the derivative of the second term, 2 pi RH. Now, we have this coefficient of 2 pi, but in order to take the derivative of R&H, we are going to need to use the product rule. So we will multiply two pi with the derivative of R and H. So by the product rule, we will take the original function R and multiply this by the derivative H with respect to H. That is just going to be 1 by the power rule, and 1 multiplied by R will just be R. Then we will add the function H multiplied by the derivative of R with respect to H. The derivative of R is going to be one by the power rule, but since we took the derivative of R with respect to H, that is going to generate a DRDH term. Now what we need to do is we need to simplify the derivative. 2 i multiplied by 2R is going to give us 4IR. And we will include the DRDH term. For the second term, we will distribute 2 pi across R plus HDRDH. That is going to give us 2 pi R plus 2 HDRDH. This is going to be the simplified form of the derivative with respect to H. But now what we want to do is we want to solve for DRDH. So what we are going to do is we are going to move any term that doesn't have a DRDH factor to the left-hand side of the equation. In order to do this, we will subtract 2 piR to both sides of the equation. That will allow us to rewrite the derivative as negative to pi R equal to 4IR DRDH. Plus 2 pi HDRDH. Notice that every term in this equation has a factor of 2 pi. So what we can go ahead and do is we can further simplify this derivative by dividing every term by 2 pi. This will allow us to simplify the derivative to be negative R equal to 2 RDRDH plus HDRDH. Now notice that every term on the right-hand side of the derivative has a multiple of DRDH. What we can go ahead and do is factor a DRDH from the right-hand side of the equation. That will allow us to rewrite the derivative as negative R equal to the quantity 2 R plus H multiplied by DRDH. And finally, in order to solve for DRDH, we are going to divide both sides of the equation by 2 R plus H. That is going to give us the final derivative DRDH equal to negative R divided by the quantity to R plus H. And this is going to be the derivative of the area of the cylinder when the area is equal to 500 pi. And with that being said, the overall solution to this problem is going to be a. So, I hope this video helps you in understanding on how to approach this problem, and I will go ahead and see you all in the next video.

Surface area of a cone The lateral surface area of a cone of radius r and height h (the surface area excluding the base) is A = πr√r²+h².a. Find dr/dh for a cone with a lateral surface area of A=1500π.

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Surface area of a cone The lateral surface area of a cone of radius r and height h (the surface area excluding the base) is A = πr√r²+h².
a. Find dr/dh for a cone with a lateral surface area of A=1500π.