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The circumference of a sphere was measured to be $ 84 cm $ with a possible error of $ 0.5 cm. $

(a) Use differentials to estimate the maximum error in the calculated surface area. What is the relative error?

(b) Use differentials to estimate the maximum error in the calculated volume. What is the relative error?

(a) $=1.2 \%$

(b) $1.8 \%$

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We've got a question here that gives us the circumference of spear sphere to be measured to be 84 centimeters, the possible error 840.5 centimeters and we want to use different essentials. Estimate maximum error in the calculated surface area. What is the relative era and the R B? We want to use the same differential to estimate maximum area and calculating the volume. Then what is the relative area? They sort of start off. We want to hurt right out. Our sir comforts equation is two pi r. You know, that's equal to 84. We can calculate our radius be equal to 84/2 pi, and that will be equal to 42 over pie. All right, now, we wanted to Frenchy both sides. We'll get the differential off. The conference is equal to two pi. We are essentially we're taking the differential of this equation. We can create in equation that gives us the change of the circumference with respect to the change of area. Excuse me. With respect to the change of radius. Okay, now we know that are change in radius. We're told that the possible area is, um students were told possible change into circumference is 0.5, right? That's the amount of error we have. So we can then use that to calculate d r and say, Well, then you are is equal to five over to pot, and that will be the same thing as 1/4. All right. Now, if we use the equation for the surface area of a sphere, we know that's the same thing as as equals to four pi r squared. All right, Now, if we did the same thing here Oh, we took the the differential of the sphere. You can calculate the change in sphere. Excuse me. The change in surface area of the sphere with respect to the change in, uh, radius. Okay, we have a pie. Are D R All right? Now, if we go and plug in the values for R and D R. Number one we know that are are was 42 pi. We calculated that to be you know, r d r. We calculated to be 1/4 1 4th pie. We can calculate our calculate our, um d s to be 84 pi. Excuse me. I'm sorry. That's should be 1/4 over pie one over pi and therefore, this would be 84 over pie. And we're told that everything is in centimeters. We'll have centimeters squared. And when you actually go ahead and divide this out, it comes out to be approximately 27 centimeters squared. OK, 84 divided by pi. All right, Now, if we want to find relative error, we would, um, asked to use Excuse me, I just re read this question. We want to use a different options estimate Maximum era in the calculus surface area. What is the relative area? Okay, so our maximum area a maximum error, it's here. And then if we wanted to calculate our relative error could use the equation changes, asshole. As is approximately the same thing as the change and STDs or us. Right? So Delta s or ask me the same thing in the same ratio approximately as the D of SRS, And we know D s or s is the same thing as ate pie. R d r over s which we know it warp. I r square right R s. Here is the surface area equation rds to derivative of that when we go into by these out. We'll be left with a two. These will cancel out. Let's go to And then one are on the bottom. The two over our We have a d r. Okay, if we go on, substitute our r and D our values. You know that our d r we said Waas, What do we say? Rdr waas 1/4 pi. Yeah, we said our our waas 42 over pie. If we go ahead and divide all that out, it will come out to approximately 1/84. And if you convert that into decimal, it is approximately point you won't want to. And we can also write that it's 1.2% there. Okay, so we can see here that the that If there is a possible area five centimeters in the circumference of the sphere than the the relative area error for the area the surface area would be 1.2%. Alright Now, part B of this question were asked to use differentials to estimate the maximum error in calculated volume. And what is the relative there? Okay, so we're gonna use a similar approach going to write out our volume equation which we know is for over three pi r huge for the sphere. And if you take the derivative of that, you'll get four pi or squared D r Here we know our r squared our was 42 over pie and R D R was one over pi. If we plug all that end 42 over pie and then we have a 1/4 pi and if we go and expand all that out it comes out to be approximately 179 and which is all it centimeters and volume is cubes. We have Centimeters Cube. Okay? And then once again, we use the same. This would be our maximum error by the way volume. And then we want to use our similar approach that we did last time when we said the change in Excuse me, the delta be over V is approximately the same thing as TV over the alright TV TV over the and that would be then equation, which is four pi r squared. We are over volume equation which we know is 4/3. I are cute and then all that simplified out comes out to be three d R over our Excuse me right now, If we go and plug in three d R. Equation. Excuse me. If you plug in D R, which we know is one of four pi and then we plug in our our, which were calculated to be 42 over pie. Then we would have all of this approximately come out to be 156 which is roughly 0.18. And we can write that also as 18%. I'm sorry. 0.18 would be one too. Yeah. Oh, I'm sorry I did this wrong. This should be 0.18 and then we would have a 1.8%. All right, well, I hope that clarifies the question first. What we did is we used our circumference equation to calculate for D of our and our our and then we use the relationships between Delta s and the Excuse me Delta, the service area over surface area and D f s over surface area to calculate the maximum error. And then we were able to get the relative error and percentage, and we did the same thing for our volume. All right, well, I hope this clarifies the question, and thank you so much for watching

The University of Texas at Arlington