Why is the margin of error 40 * pi * h for this problem?

Question

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 $40\pi$.

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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.

Answer 1

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 $\dfrac{\pi}{4}\left(6^2+x^2\right)$.

Answer 2

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 $40\pi$.