Volume of a sphere.

: Spherical balloon

Step 1: Volume of a sphere.
$V = \frac{4}{3} \pi r^3$

Step 2: Differentiate both sides with respect to time $t$ (chain rule).
$\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$

Step 3: 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

Step 1: Pythagorean theorem for ladder length $l = 10$ m (constant).
$x^2 + y^2 = 10^2 = 100$

Step 2: Differentiate both sides with respect to time $t$.
$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$

Step 3: Solve for $\frac{dy}{dt}$.
$\frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt}$

Step 4: Find $y$ when $x = 6$ m.
$y = \sqrt{100 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8$ m

Step 5: 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}}