How do you set up a related rates problem

Related rates problems ask how one changing quantity influences another when both vary with time. The first task is to draw a clear diagram of the situation and label every length, angle, or area that appears. Next, write an equation that connects these quantities using geometry or physics, such as the Pythagorean theorem for a right‑triangle.

Step by step procedure

After the equation is in place, treat every variable as a function of time $t$. Differentiate both sides with respect to $t$ using implicit differentiation, remembering to multiply each derivative by the corresponding $\frac{dx}{dt}$ term. Finally, plug in the numerical values for the known rates at the instant of interest and solve the resulting algebraic equation for the unknown rate.

Key decisions before differentiating:

• Which variable is the one you need to find?
• Which variables have known rates at the instant of interest?
• What units are each quantity measured in?

Consider a 10‑ft ladder leaning against a vertical wall. The foot of the ladder slides away from the wall at $3\text{ ft/s}$. When the foot is 6 ft from the wall, find the speed at which the top slides down. Let $x$ be the distance from the wall to the foot, $y$ the height of the top, and the ladder length $L=10$ ft. The relationship is $x^{2}+y^{2}=L^{2}$. Differentiate: $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$. Plug in $x=6$, $\frac{dx}{dt}=3$, and solve for $\frac{dy}{dt}$. First compute $y=\sqrt{L^{2}-x^{2}}=\sqrt{100-36}=8$ ft. Then $2(6)(3)+2(8)\frac{dy}{dt}=0$ gives $36+16\frac{dy}{dt}=0$, so $\frac{dy}{dt}=-\frac{36}{16}=-2.25\text{ ft/s}$. The negative sign indicates the top is descending.

Applying the method to the example:

1. 1Write the geometric relation $x^{2}+y^{2}=10^{2}$.
2. 2Differentiate to obtain $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$.
3. 3Insert $x=6$, $\frac{dx}{dt}=3$, and $y=8$.
4. 4Solve for $\frac{dy}{dt}$ to get $-2.25$ ft/s.

Always check that the units of the computed rate match the physical interpretation. If the answer is in feet per second but the problem describes a volume change, convert to cubic feet per second before concluding. A quick unit check often catches algebraic sign errors before the test.