Question

a sphere is to be designed with a radius of 72 in. use differentials to estimate the maximum error when measuring the volume of the sphere if the possible error in measuring the radius is 0.5 in. (hint: the formula for the volume of a sphere is $v(r)=\frac{4}{3}pi r^3$.) 32,572.03 in³ 65,144.07 in³ 452.39 in³ 16,286.02 in³

Explanation:

Step1: Differentiate volume formula

The volume of a sphere is $V(r)=\frac{4}{3}\pi r^3$. Take the derivative with respect to $r$:
$$\frac{dV}{dr}=4\pi r^2$$
Rewrite as $dV=4\pi r^2 dr$

Step2: Substitute given values

We know $r=72$ in, $dr=0.5$ in. Substitute into the differential:
$$dV=4\pi (72)^2 (0.5)$$

Step3: Calculate the result

First compute $(72)^2=5184$, then:
$$dV=4\pi \times 5184 \times 0.5=2\pi \times 5184=10368\pi$$
Calculate the numerical value: $10368\pi \approx 32572.03$

Answer:

32,572.03 $\text{in}^3$