Related+Rates

=__Related Rates__=

One of the most prominent applications of Implicit Differentiation is solving problems where multiple are changing in relation to each other, commonly called Related Rates problems. For example, if the bottom of a ladder in being pulled away from a wall, the rate at which it is being pulled will directly affect the rate the top of the ladder slides down the wall.

Most commonly, you will be given one constant rate and a some of the conditions at the time in question, and will be asked to find the rest of the conditions at that time. You will be called on to remember many formulas to relate the variables to each other, most often the Pythagorean Theorem and Volume formulas.

The following examples showcase the most common Related Rates problems.

**Example One** The bottom of a 10 foot ladder is being pulled away from a wall at a rate of 1 foot/minute. How fast is the top of the ladder sliding down the wall when it is 5 feet from the ground?

Solution: The wall, ground and ladder form right triangle, so the Pythagorean Theorem can be applied: **x 2 +y 2 =(10) 2 **, where x is the distance of the ladder's bottom from the wall and y is the distance from the ladder's top to the ground. We are given that the rate x is changing with respect to time, **dx/dt, is 1 ft/min** and that **y is 5 ft**. Using y=5, we can plug into the above equation to find that **x=sqrt(75)=5sqrt(3****) ft**. The problem asks us to find dy/dt, so we will now differentiate implicitly and solve for dy/dt. The ladder is sliding down the wall at **sqrt(3) ft/min**.

**Example Two** A 6 ft tall person walks away from a light 12 ft off the ground at a rate of 3 ft/sec. How fast is the tip of the shadow going away from the light when he is 8 ft from the light?

Solution: Refer to the diagram to the right. We will define T=s+d We are given that dd/dt=3 ft/sec, and we want dT/dt when d=8 ft. Now let's set up similar triangles, which yields 12/(d+s)=6/s Rearranging, we get 12s=6d+6s, and that s=d Substituting, we get that T=2d Take the Derivative yielding: dT/dt=2dd/dt=2(3 ft/sec)=**6 ft/sec**.

**Example Three** An airplane is flying directly towards an observer on the ground at an altitude of 5000 feet at a horizontal speed of 500 feet/second. How fast is the distance between the plane and observer changing the observation angle from the ground to the plane is (p ﻿/6) radians?

Solution: sin( p /6)=a/d and tan( p /6)=a/h, where a=altitude, h=horizontal distance, d=diagonal distance. Using these relations with a=5000 ft, we find that h=5000sqrt(3) ft and d=10000 ft We can now set up a right triangle, apply the Pythagorean Theorem, and differentiate implicitly to find the desired dd/dt:

**Example Four** A perfectly spherical balloon is inflated at a rate of (1/2) cubic meters per minute. What is the diameter of the balloon when the diameter of the balloon is increasing at (p ) m/min? (adapted from the 2011 St. Valentine's Day Mathacre Varsity Test, written by Zachary Youngblood)

Solution: We are given dV/dt=(1/2) m 3 /min and dd/dt= (p ) m/min.

**Example Five** A tank that is in the shape of a right circular cone is being filled at a constant rate of 5 m 3 /min. The radius of the tank at its base is 5 m and the height of the tank is 10 m. When the depth of the water in the tank is 6 m, how fast is the height of the tank rising?

Solution: We are given that dV/dt=5 m 3 /min, that h=6 m, and the dimensions of the tank. We want to find dh/dt. Refer to the diagram below to set up similar triangles.

By similar triangles we have: (r/h)=(5/10), yielding r=(1/2)h.

We are now ready to differentiate the formula for the Volume of a Cone by getting Volume in terms of h only: