What if the given rate is a decrease?

Assign a negative sign to the rate. For example, if volume is leaking at 5 L/min, $\frac{dV}{dt} = -5$.

Can I use similar triangles to eliminate a variable?

Yes. In problems like conical tanks or shadows, you often have more variables than known rates. Use geometric relationships to express one variable in terms of another before differentiating.

When should I use trigonometric ratios?

Use them when the problem involves an angle changing over time or asks for the rate of change of an angle. The governing equation will be a trig function like $ an( heta) = \frac{y}{x}$.

How to solve a related rates problem

Related rates problems require finding the rate of change of one quantity by relating it to other quantities whose rates of change are known. This method applies when multiple variables depend on time and are connected by a geometric or physical equation.

The procedure uses implicit differentiation with respect to time $t$ . By applying the chain rule to the governing equation, you can substitute the known values and rates to solve for the unknown rate.

The setup

Identify all given quantities, known rates of change, and the unknown rate of change to be found. Define variables for all quantities that change over time, and assign constant values only to quantities that do not change.

The steps

Draw a diagram and label all variables and constants. 2. Write an equation relating the variables (e.g., Pythagorean theorem, volume formulas, trigonometry). 3. Differentiate both sides of the equation implicitly with respect to time $t$ . Every variable $x$ differentiated with respect to $t$ yields $\frac{dx}{dt}$ . 4. Substitute all known values and known rates of change into the differentiated equation. 5. Solve algebraically for the unknown rate of change.

Checking the result

Verify that the sign of the computed rate makes physical sense. If a quantity is decreasing, its rate of change must be negative. Ensure the units of the final answer match the units of the differentiated variable with respect to time.

Common errors

Substituting known variable values before differentiating. This eliminates variables prematurely and results in zero derivatives. Another error is forgetting to apply the chain rule, omitting the $\frac{dx}{dt}$ terms.

Worked example

A 10-foot ladder is leaning against a vertical wall. The bottom of the ladder is sliding away from the base of the wall at a rate of 2 ft/sec. How fast is the top of the ladder sliding down the wall when the bottom is 6 feet from the base?

Let $x$ be the distance from the bottom of the ladder to the wall, and $y$ be the distance from the top of the ladder to the ground. Let $L=10$ be the length of the ladder. We are given $\frac{dx}{dt} = 2$ ft/sec. We need to find $\frac{dy}{dt}$ when $x = 6$ .

The variables are related by the Pythagorean theorem: $x^2 + y^2 = 10^2$

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

Find the value of $y$ when $x = 6$ : $6^2 + y^2 = 100$ $36 + y^2 = 100$ $y^2 = 64$ $y = 8$

Substitute $x = 6$ , $y = 8$ , and $\frac{dx}{dt} = 2$ into the derivative equation: $(6)(2) + (8)\frac{dy}{dt} = 0$ $12 + 8\frac{dy}{dt} = 0$ $8\frac{dy}{dt} = -12$ $\frac{dy}{dt} = -1.5$

The top of the ladder is sliding down the wall at a rate of 1.5 ft/sec.

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References: Calculus, Early Transcendentals by James Stewart · OpenStax Calculus Volume 1, Chapter 4.1