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For my Calculus assignment, I was given this problem:

  1. If a right triangle has legs 6 and 8, its hypotenuse is 10. The triangle will be inscribed within a circle with area 25pi. (The hypotenuse will be the diameter of the circle).

A. Suppose one leg of the triangle is known to be exactly 6, but the other leg is known to be 8 with an error of +-h. What are x, f(x), and a in this problem?

^ the above question about x, f(x), and a, are regarding the format that was taught for solving Tangent Line Approximation problems.

C. Now consider the sphere that just contains the triangle (so the hypotenuse is the diameter of the sphere). Use a tangent line approximation to estimate the volume of this sphere.

For the answer, I ended up with $\frac{500\pi}{3}\pm 100\pi h$

My teacher said the margin of error was only $40\pi h$

I don't quite understand why though...

The Tangent Line Approximation form we're given is: $f(x) + f'(x)(a - x)$, where x is the known value, f' is the derivative of f, and a is the value being solved for.

For the volume of a sphere, I have $V = \frac{4}{3}\pi r^3$

For the radius of the sphere, I used $\frac{\sqrt{x^2+36}}{2}$

I found the volume of a triangle with the side measuring 8 to be $\frac{500 \pi}{3}$ and its derivative: $100 \pi$

Plugging everything into the formula: $\frac{500 \pi}{3}+100 \pi(8\pm h-8) = \frac{500 \pi}{3} \pm 100 \pi h$

I see that if the derivative was $40\pi$, then the margin of error would be what my teacher had said, but I don't see how it makes sense otherwise, or why the derivative would be $40\pi$.


I'm slow with LaTex, so I will post the question and add the work I've done for this problem so far while it's being looked at.

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2 Answers 2

up vote 3 down vote accepted

For the sphere problem, if $x$ is the length of the side which is nearly $8$, the volume $V(x)$ of the sphere is $\dfrac{4\pi}{3}r^3$, where $r(x)=\dfrac{\sqrt{36+x^2}}{2}$.

Substituting in the formula for the volume of the sphere, and simplifying a bit, we obtain $$V(x)=\frac{\pi}{6}(36+x^2)^{3/2}.$$ We want to use the tangent line approximation to approximate $V(x)$ when $x$ is near $8$. Differentiate. We get $$V'(x)=\pi\frac{x}{2}(36+x^2)^{1/2}.$$ Here the Chain Rule was used.) Set $x=8$. At $x=8$, the derivative is $40\pi$.

The tangent line approximation says that $$V(a)\approx V(8)+40\pi(x-8).$$ If we know that "$x=8\pm h$," where $h$ is positive and small, then by the tangent line approximation we have approximated $V(8)$ within roughly $\pm 40\pi h$.

The area problem is technically easier, since the area is given by the simple expression $\dfrac{\pi}{4}\left(6^2+x^2\right)$.

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Where's pi in the volume formula, and is the equation with r(x) supposed to be over two? Sorry, I am confused. –  mowwwalker Dec 8 '12 at 7:09
1  
Sorry, lost my $\pi$'s. I have a habit of leaving out such constants and putting them in at the end. Except forgot. Thanks. I think they are all back. –  André Nicolas Dec 8 '12 at 7:15
    
I mixed up my a's and x's in the formula - what you posted is, if I remember correctly, what was taught. Also, in your formula for radius, it needs to be over 2, but right now the 2 is just to the right of the equation. –  mowwwalker Dec 8 '12 at 7:35
    
@Walkerneo: I am relieved! With the opposite notation there would have been much confusion in subsequent courses. Have removed comment. Have fixed TeX error that misplaced the $2$. Hope that there are no more typos. –  André Nicolas Dec 8 '12 at 7:42

The problem I had was that I was finding the derivative of the volume function, V, in terms of the triangle's side, x, but didn't apply the chain rule to the function I used for the radius, r.

The derivative I had for the volume was $4\pi r^2$, but it should have been $4\pi r^2 r'$

The derivative of $r$ would be $\frac{2}{5}$, which, when multiplied by the derivative I did get ($100\pi$) would have landed me with $40\pi$.

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