The volume of a sphere is increasing at $8\pi$ cm³/s. The rate of change of surface area when the radius is 2 cm is:
A$2\pi$ cm²/s
B$4\pi$ cm²/s
C$8\pi$ cm²/s
D$16\pi$ cm²/s
Answer & Solution
Correct answer: C. $8\pi$ cm²/s
$V = (4/3)\pi r^3$ ⇒ $dV/dt = 4\pi r^2 \cdot dr/dt = 8\pi$.
At $r = 2$: $4\pi(4)\cdot dr/dt = 8\pi$ ⇒ $dr/dt = 1/2$.
$S = 4\pi r^2$ ⇒ $dS/dt = 8\pi r \cdot dr/dt = 8\pi(2)(1/2) = 8\pi$ cm²/s.
Keep practising Unit 4-5: Applications of Derivatives & Curve Sketching
Browse all Unit 4-5: Applications of Derivatives & Curve Sketching questions for AP Calculus AB, or explore more Mathematics topics.
Related questions
If the tangent to the curve $y = e^{2x}$ at point $(a, e^{2a})$ passes through origin, theIf $f(x) = x^x$ for $x > 0$, then $f$ has a minimum at $x = $:The function $f(x) = \dfrac{x}{1 + x^2}$ attains its maximum value on $[0, \infty)$ at:Find the **shortest distance** from origin to the curve $xy = 4$ (in the first quadrant):A rectangle is inscribed in a semicircle of radius $R$ with base on the diameter. The **maThe function $f(x) = x^3 - 3x + 1$ has critical points at:$f(x) = 2x^3 - 3x^2 - 12x + 5$. The **local maximum value** occurs at:Find the approximate value of $ qrt{25.5}$ using differentials (given $ qrt{25} = 5$):