4.5 Solving Related Rates Problems

Learning objectives

—CHA-3.E—Interpret related rates in applied contexts• CHA-3.E Interpret related rates in applied contexts.• CHA-3.E.1 The derivative can be used to solve related rates problems; that is, finding a rate at which one quantity is changing by relating it to other quantities whose rates of change are known.• Enduring understanding CHA-3: Derivatives allow us to solve real-world problems involving rates of change.

To solve a related-rates problem, connect the changing quantities with one equation, differentiate with respect to their common variable, and use known values from the same instant to determine the requested rate.

A 10-foot ladder rests against a wall. Let $x$ be the bottom's distance from the wall and $y$ the top's height, both in feet. The fixed length gives $x^2+y^2=100$ . Suppose the bottom moves away at $dx/dt=2$ feet per second. Find the top's rate when $x=6$ .

2x\frac{dx}{dt}+2y\frac{dy}{dt}=0

At the specified instant, $6^2+y^2=100$ , so $y=8$ feet. Substitute $x=6$ , $y=8$ , and $dx/dt=2$ : $2(6)(2)+2(8)\frac{dy}{dt}=0,$ hence $\frac{dy}{dt}=-\frac{24}{16}=-1.5\ \mathrm{ft/s}.$ When the bottom is 6 feet from the wall, the top is moving downward at 1.5 feet per second.

The negative result is not an error: with upward chosen as positive, it means the height is decreasing. Use positions and rates from the same instant, and state both direction and units; writing only $-1.5$ leaves the contextual answer incomplete.

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—4.6 Approximating Values of a Function Using Local Linearity and LinearizationRead