Related rates is just an application of the chain rule. Typically, we have a function y=x, where x is implicitly dependent on another variable, usually time, t. Thus, if we differentiate the equation with respect to time, we have dydt=dydxdxdt. Related rates problems usually give su two of these differentials, and ask us to compute the third.
Example 1
Taken from practice exercise in Cracking the GRE Mathematics Subject Test, 4th EditionThe radius of a circle is decreasing at a rate of 0.5 cm per second. At what rate in cm2/sec is the circle's area decreasing when the radius is 4 cm?
The area of the circle is given by A=πr2. Differentiating with respect to time, t, we have
dAdt=2πrdrdtWe are given that drdt=−0.5, and that, at the refrence time t0, the radius is 4. Plugging these values in, we have
dAdt=2π(4)(−0.5)=−4πThus, the area is decreasing by −4π cm2/sec.
Example 2
Taken from practice exercise in Calculus with Analytic Geometry, Third EditionA board 5 feet long slides down a wall. At the instant the bottom end is 4 feet from the wall, the other end is moving down the wall at the rate of 2 feet per second. At that moment, how fast is the bottom end sliding along the ground?
If we let y denote the height of the board against the wall, and x the distance of the board from the wall, then we have the equation
52=x2+y2Differentiating with respect to time, we have
0=2xdxdt+2ydydtWe are given that, at the reference time, t0, dxdy=−2 and x=4. Since, x=4, we know that y=3 (it's a 3-4-5 triangle). Plugging these values into the above equation, we have
0=2(4)dxdt+2(3)(−2)Which gives us dxdt=32. Thus, the bottom of the board is sliding at a rate of 1.5 feet per second.