Related Rates

$$V=\frac{4}{3}\pi r^3,\quad S=4\pi r^2$$

The balloon is modeled as a perfect sphere. Both V and r depend on time.

$$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt},\quad\frac{dS}{dt}=8\pi r\frac{dr}{dt}$$

The chain rule relates both rates to dr/dt. The illustrative inflow of one cubic inch per 20 seconds is dV/dt=1/20 in³/s; deflation at the same magnitude has the negative rate.

$$S=4\pi\left(\frac{3V}{4\pi}\right)^{2/3}$$

Eliminating r relates surface area directly to volume. The exponent 2/3 comes from squaring the cube root; physically r≥0 and V≥0.

$$R=10\ \mathrm{ft},\quad H=6\ \mathrm{ft},\quad\frac{r}{h}=\frac{10}{6}$$

The container dimensions stay fixed. Similar triangles relate the water radius to its height.

$$r=\frac{5}{3}h,\qquad\frac{dr}{dt}=\frac{5}{3}\frac{dh}{dt}$$

The water dimensions change together. This is a model-building illustration rather than a completed numerical inflow problem.

$$\frac{dL}{dt}=2\ \mathrm{in/s},\quad\frac{dw}{dt}=3\ \mathrm{in/s}$$

The stated instant has L=10 inches and w=6 inches.

$$P=2L+2w\quad\Longrightarrow\quad\frac{dP}{dt}=2\frac{dL}{dt}+2\frac{dw}{dt}$$

Differentiate the perimeter relationship.

$$\frac{dP}{dt}=2(2)+2(3)=10\ \mathrm{in/s}$$

The dimensions do not appear in the rate equation. If the two given dimension rates remain constant, this perimeter rate remains constant.

$$L=10,\quad w=6,\quad\frac{dL}{dt}=3,\quad\frac{dw}{dt}=-2$$

Lengths are inches and rates are inches per second. This question asks for area, so use a new model.

$$A=Lw\quad\Longrightarrow\quad\frac{dA}{dt}=w\frac{dL}{dt}+L\frac{dw}{dt}$$

Use the product rule before substituting the dimensions.

$$\frac{dA}{dt}=6(3)+10(-2)=-2\ \mathrm{in^2/s}$$

The area decreases at 2 square inches per second. A length can grow while the total area shrinks.

$$r=20\ \mathrm{cm},\quad\frac{dr}{dt}=-3\ \mathrm{cm/min}$$

The radius is decreasing at the given instant.

$$V=\frac{4}{3}\pi r^3\quad\Longrightarrow\quad\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$$

Differentiate with respect to time before substituting r=20.

$$\frac{dV}{dt}=4\pi(20)^2(-3)=-4800\pi\ \mathrm{cm^3/min}$$

The exact result is negative; the volume decreases at 4800π cubic centimeters per minute.

$$\frac{dV}{dt}\approx-15079.645\ \mathrm{cm^3/min}$$

This is an approximation rounded to three decimal places.

$$h^2+b^2=25^2,\quad\frac{db}{dt}=3\ \mathrm{ft/s}$$

Let b be the base’s distance from the wall and h be the top’s height. The ladder length is fixed.

$$2h\frac{dh}{dt}+2b\frac{db}{dt}=0\quad\Longrightarrow\quad\frac{dh}{dt}=-\frac{b}{h}\frac{db}{dt}$$

The derivative of the fixed 25² is zero.

$$b=20\ \Longrightarrow\ h=\sqrt{25^2-20^2}=15\ \mathrm{ft}$$

Find the missing height at this instant.

$$\frac{dh}{dt}=-\frac{20}{15}(3)=-4\ \mathrm{ft/s}$$

The signed height rate is −4 ft/s. The top moves downward at a speed of 4 ft/s.

$$\sin\theta=\frac{b}{25}\quad\Longrightarrow\quad\cos\theta\frac{d\theta}{dt}=\frac{1}{25}\frac{db}{dt}$$

Place θ between the ladder and the vertical wall, rather than between the ladder and the ground.

$$b=15\ \Longrightarrow\ h=20,\quad\cos\theta=\frac{20}{25}=\frac{4}{5}$$

This is a different instant from the previous question.

$$\frac{d\theta}{dt}=\frac{3}{25(4/5)}=\frac{3}{20}\ \mathrm{rad/s}$$

The positive result agrees with the wall angle increasing as the base moves outward. The ground angle decreases at the same magnitude.