Problem 12: Spherical balloon Step 1: Volume of a sphere. V = 4/3 π r³ Step 2: Differentiate both sides with respect to time t (chain rule). dV/dt = 4 π r² dr/dt; Step 3: Substitute r = 5 cm and dr/dt = 2 cm/s. dV/dt = 4 π (5)² (2) = 4 π (25)(2) = 200 π; dV/dt = 200π cm³/s 200π Problem 13: Ladder problem Step 1: Pythagorean theorem for ladder length l = 10 m (constant). x² + y² = 10² = 100 Step 2: Differentiate both sides with respect to time t. 2x dx/dt + 2y dy/dt = 0 Step 3: Solve for dy/dt. dy/dt = −x/y dx/dt; Step 4: Find y when x = 6 m. y = √(100 − 6²) = √(100 − 36) = √(64) = 8 m; Step 5: Substitute x = 6 m, y = 8 m, dx/dt = 1 m/s. dy/dt = −6/8 (1) = −3/4 = −0.75 m/s The top is sliding down at 0.75 m/s. dy/dt = −3/4 m/s −3/4

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Volume of a sphere.

March 18, 2026Mathematics

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Answer

200\pi

-\frac{3}{4}

Step-by-step solution

8 steps

: Spherical balloon

Volume of a sphere.
$V = \frac{4}{3} \pi r^3$
Differentiate both sides with respect to time $t$ (chain rule).
$\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$
Substitute $r = 5$ cm and $\frac{dr}{dt} = 2$ cm/s.
$\frac{dV}{dt} = 4 \pi (5)^2 (2) = 4 \pi (25)(2) = 200 \pi$
$\frac{dV}{dt} = 200\pi$ cm³/s

200\pi

: Ladder problem
Pythagorean theorem for ladder length $l = 10$ m (constant).
$x^2 + y^2 = 10^2 = 100$
Differentiate both sides with respect to time $t$ .
$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$
Solve for $\frac{dy}{dt}$ .
$\frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt}$
Find $y$ when $x = 6$ m.
$y = \sqrt{100 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8$ m
Substitute $x = 6$ m, $y = 8$ m, $\frac{dx}{dt} = 1$ m/s.
$\frac{dy}{dt} = -\frac{6}{8} (1) = -\frac{3}{4} = -0.75$ m/s

The top is sliding down at 0.75 m/s.

$\frac{dy}{dt} = -\frac{3}{4}$ m/s

-\frac{3}{4}

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Volume of a sphere.

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Step-by-step solution