Question

Approximate volume of a cone whose semi-vertical angle is \[ \tan^{-1}(3) \] and base radius is \(15.001\) is

• A. \((375.025)\pi\)
• B. \((325.025)\pi\)
• C. \((375.075)\pi\)
• D. \((325.075)\pi\)

Hint:

For approximation problems, use \[ \boxed{f(x+\Delta x)\approx f(x)+f'(x)\Delta x.} \] Here, \[ V=\frac{\pi r^3}{9}, \] so \[ dV=\frac{\pi}{3}r^2\,dr. \]

Answer 1

The Correct Option is C

Step 1: Find the height of the cone. The semi-vertical angle satisfies \[ \tan\theta=\frac{r}{h}=3. \] Hence, \[ h=\frac{r}{3}. \] The volume is \[ V = \frac13\pi r^2h = \frac{\pi r^3}{9}. \]

Step 2: Use differential approximation. Let \[ V=\frac{\pi r^3}{9}. \] Then, \[ dV = \frac{\pi}{3}r^2\,dr. \] Here, \[ r=15,\qquad dr=0.001. \] Therefore, \[ dV = \frac{\pi}{3}(15)^2(0.001) = 0.075\pi. \]

Step 3: Compute the approximate volume. At \[ r=15, \] \[ V = \frac{\pi(15)^3}{9} = 375\pi. \] Hence, \[ V+dV = 375\pi+0.075\pi = (375.075)\pi. \] Thus, \[ \boxed{(375.075)\pi.} \] Therefore, the correct option is \(\boxed{(C)}\).