11DIFFERENTIATION - Department of Mathematics
11DIFFERENTIATION - Department of Mathematics
11DIFFERENTIATION - Department of Mathematics
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766 11 DIFFERENTIATION<br />
Solving Related Rates<br />
Problems<br />
EXAMPLE 8<br />
FIGURE 11.17<br />
The rate at which x is changing with respect<br />
to time is related to the rate <strong>of</strong> change <strong>of</strong><br />
y with respect to time.<br />
1. Assign a variable to each quantity. Draw a diagram if needed.<br />
2. Write the given values <strong>of</strong> the variables and their rates <strong>of</strong> change with respect<br />
to t.<br />
3. Find an equation giving the relationship between the variables.<br />
4. Differentiate both sides <strong>of</strong> this equation implicitly with respect to t.<br />
5. Replace the variables and their derivatives by the numerical data found in<br />
step 2 and solve the equation for the required rate <strong>of</strong> change.<br />
At a distance <strong>of</strong> 4000 feet from the launch site, a spectator is observing a<br />
rocket being launched. If the rocket lifts <strong>of</strong>f vertically and is rising at a speed<br />
<strong>of</strong> 600 feet/second when it is at an altitude <strong>of</strong> 3000 feet, how fast is the distance<br />
between the rocket and the spectator changing at that instant?<br />
SOLUTION ✔ Step 1 Let<br />
y the altitude <strong>of</strong> the rocket<br />
x the distance between the rocket and the spectator<br />
at any time t (Figure 11.17).<br />
x<br />
4000 ft<br />
Rocket<br />
Step 2 We are given that at a certain instant <strong>of</strong> time<br />
y 3000 and<br />
dy<br />
600<br />
dt<br />
Step 3<br />
and are asked to find dx/dt at that instant.<br />
Applying the Pythagorean theorem to the right triangle in Figure<br />
11.17, we find that<br />
x2 y2 40002 Therefore, when y 3000,<br />
y<br />
Launching pad<br />
x 3000 2 4000 2 5000