Calculus Reference
Related Rates
Related rates connect time derivatives of quantities through an equation that describes their relationship.
Solved Examples
Examples cover a changing distance constraint and a changing volume.
Time Derivative Formulas
| Quantity | Time derivative |
|---|---|
| $x^2$ | $2x\frac{dx}{dt}$ |
| $xy$ | $x\frac{dy}{dt}+y\frac{dx}{dt}$ |
| $\frac43\pi r^3$ | $4\pi r^2\frac{dr}{dt}$ |
Rates carry units per unit of time. A negative rate means the corresponding quantity is decreasing.
Sliding Ladder
10-Foot Ladder
The bottom moves away from a wall at $3$ ft/s. Find the height's rate of change when the bottom is $6$ ft from the wall.
Solution Steps
- Relate the bottom distance $x$ and height $y$ using the fixed ladder length.
- Differentiate the relation with respect to time.
- Solve the differentiated relation for $dy/dt$.
- Substitute the dimensions and known rate.
$$x^2+y^2=100,\qquad y=8$$$$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$$$$\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$$$$\boxed{\frac{dy}{dt}=-\frac{6}{8}(3)=-\frac94\text{ ft/s}}$$
Growing Sphere
Volume Growth
A sphere's radius increases at $2$ cm/s. Find the volume's rate of change when the radius is $3$ cm.
Solution Steps
- Write the volume formula in terms of the radius.
- Differentiate the volume formula with respect to time.
- Substitute the radius and its rate of change.
$$V=\frac43\pi r^3$$$$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$$$$\boxed{\frac{dV}{dt}=4\pi(3)^2(2)=72\pi\text{ cm}^3/\text{s}}$$