Derivative applications turn formulas into real-world models. In related rates, several quantities change with
time. In optimization, a derivative helps find maximum or minimum values.
1. Related rates idea
Related rates problems connect two or more changing quantities. The general strategy is:
- Draw or describe the physical situation.
- Write an equation connecting the variables.
- Differentiate both sides with respect to time \(t\).
- Substitute known values after differentiating.
- Solve for the unknown rate.
The chain rule is the key idea:
\[
\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}
\]
2. Ladder sliding problem
A ladder of length \(L\) leans against a wall. Let \(x\) be the distance of the bottom from the wall and
\(y\) be the height of the top.
\[
x^2+y^2=L^2
\]
Differentiate with respect to time:
\[
2x\frac{dx}{dt}+2y\frac{dy}{dt}=0
\]
Solve for the vertical rate:
\[
\boxed{
\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}
}
\]
The negative sign means that if the bottom moves away from the wall, the top moves downward.
3. Balloon volume rate
For a spherical balloon:
\[
V=\frac{4}{3}\pi r^3
\]
Differentiate with respect to time:
\[
\boxed{
\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}
}
\]
This tells us how fast the volume changes when the radius changes.
4. Cone water level problem
For a cone with tank radius \(R\), tank height \(H\), and water height \(h\), similar triangles give:
\[
\frac{r}{h}=\frac{R}{H}
\]
\[
r=\frac{R}{H}h
\]
Substitute into the cone volume formula:
\[
V=\frac{1}{3}\pi r^2h
\]
\[
V=\frac{\pi R^2}{3H^2}h^3
\]
Differentiate:
\[
\boxed{
\frac{dV}{dt}=\frac{\pi R^2}{H^2}h^2\frac{dh}{dt}
}
\]
5. Shadow problem
A person of height \(h\) walks away from a lamp of height \(H\). Let \(x\) be the distance from the lamp and
\(s\) be the length of the shadow.
\[
\frac{H}{x+s}=\frac{h}{s}
\]
Solving for \(s\):
\[
s=\frac{hx}{H-h}
\]
Differentiate:
\[
\boxed{
\frac{ds}{dt}=\frac{h}{H-h}\frac{dx}{dt}
}
\]
The tip of the shadow moves at:
\[
\boxed{
\frac{d(x+s)}{dt}=\frac{H}{H-h}\frac{dx}{dt}
}
\]
6. Optimization preview
Optimization problems use derivatives to find maximum or minimum values.
For a rectangle with fixed perimeter \(P\):
\[
2x+2y=P
\]
\[
y=\frac{P}{2}-x
\]
Area is:
\[
A=xy
\]
\[
A(x)=x\left(\frac{P}{2}-x\right)
\]
Differentiate:
\[
A'(x)=\frac{P}{2}-2x
\]
Set the derivative equal to zero:
\[
\frac{P}{2}-2x=0
\]
\[
\boxed{x=\frac{P}{4}}
\]
The maximum area occurs when the rectangle is a square:
\[
\boxed{x=y=\frac{P}{4}}
\]
7. Related rates vs. optimization
| Topic |
Main question |
Typical derivative |
Example |
| Related rates |
How fast is something changing? |
\(\dfrac{dy}{dt}\) |
Ladder, shadow, balloon, cone. |
| Optimization |
What value makes something largest or smallest? |
\(\dfrac{dA}{dx}=0\) |
Maximum area or minimum cost. |
| Linear approximation |
How can a tangent line approximate nearby values? |
\(L(x)=f(a)+f'(a)(x-a)\) |
Approximate square roots or measurements. |
8. Common mistakes
- Substituting numbers before differentiating.
- Forgetting that variables depend on time in related rates problems.
- Using the wrong sign for a rate. Moving downward or shrinking should usually be negative.
- Forgetting to use similar triangles in cone and shadow problems.
- Finding \(A'(x)=0\) in optimization but forgetting to check whether it gives a maximum or minimum.
- Mixing units, such as meters and centimeters, in the same calculation.
9. Modeling checklist
- Define every variable clearly.
- Write the equation before differentiating.
- Differentiate with respect to the correct independent variable.
- Substitute values only after differentiating.
- Attach units to the final answer.
- Interpret the sign of the result.
