Related+Rates

**USEFUL EQUATIONS AND FACTS TO KNOW**

 * Cones**
 * Equations**
 * Volume = 1/3 Pi * R^2** * H
 * Surface Area = Pi *R^2 + Pi * R of the Base**
 * Cubes**
 * Equations**
 * Volume = Any side^3**
 * Surface Area = 6 side^2**
 * Relations**
 * Surface Area = Twice the Derivative of Volume**
 * Volume = Half the Integral of Surface Area**


 * Cylinders**
 * Equations**
 * Volume= H * Pi R^2**
 * Surface Area= 2PiR^2 + H * 2PiR**
 * Pyramids**
 * V=(1/3)LWH**
 * Spheres**
 * SA=4PiR^2**
 * V=(4/3)PiR^3**
 * SA is the derivative of Volume**
 * Usually when solving a questions involving related rates they will ask you to find the rate of change for a certain quantity and I always start by deriving whatever equation fit the problem to find the eqution for the answer. The look and see what variable the equation is based on and find that which is usually given somewhere in the question with everything else you need.**
 * Examples Questions from __//pauls online math notes//__**
 * //Example 1//** Air is being pumped into a spherical balloon at a rate of 5 cm3/min. Determine the rate at which the radius of the balloon is increasing when the diameter of the balloon is 20 cm.

The first thing that we’ll need to do here is to identify what information that we’ve been given and what we want to find. Before we do that let’s notice that both the volume of the balloon and the radius of the balloon will vary with time and so are really functions of time, V(t) and r(t) We know that air is being pumped into the balloon at a rate of 5 cm3/min. This is the rate at which the volume is increasing. Recall that rates of change are nothing more than derivatives and so we know that, V prime=5 We want to determine the rate at which the radius is changing. Again, rates are derivatives and so it looks like we want to determine, r prime(t)=? whem r(t)=d/2=10cm Now that we’ve identified what we have been given and what we want to find we need to relate these two quantities to each other. In this case we can relate the volume and the radius with the formula for the volume of a sphere. V(t)=4/3Pi r(t)^3 As in the previous section when we looked at implicit differentiation, we will typically not use the (t) part of things in the formulas, but since this is the first time through one of these we will do that to remind ourselves that they are really functions of //t//. Now we don’t really want a relationship between the volume and the radius. What we really want is a relationship between their derivatives. We can do this by differentiating both sides with respect to //t//. In other words, we will need to do implicit differentiation on the above formula. Doing this gives, dV/dT=4Pi r^2 dr/dt Note that at this point we went ahead and dropped the (t) from each of the terms. Now all that we need to do is plug in what we know and solve for what we want to find. 5 = 4Pi (10^2) dr/dt dr/dt=1/80Pi cm/min
 * //Solution//**

How fast is the top of the ladder moving up the wall 12 seconds after we start pushing?
 * //Example 2//** A 15 foot ladder is resting against the wall. The bottom is initially 10 feet away from the wall and is being pushed towards the wall at a rate of 1/4 ft/sec

.1319 ft/sec

At what rate is distance between the two people changing when the angle is .5 radians
 * //Example 3//** Two people are 50 feet apart. One of them starts walking north at a rate so that the angle shown in the diagram below is changing at a constant rate of.01 rad/min

.311254 ft/sec

. The base radius of the tank is 5 ft and the height of the tank is 14 ft.
 * //Example 4//** A tank of water in the shape of a cone is leaking water at a constant rate of 2 ft^3 / Hour
 * (a)** At what rate is the depth of the water in the tank changing when the depth of the water is 6 ft?
 * (b)** At what rate is the radius of the top of the water in the tank changing when the depth of the water is 6 ft?

-.1386 ft/hr -.04951 ft/hr

At what rate is the height of the water changing when the water has a height of 120 cm?
 * //Example 5//** A trough of water is 8 meters deep and its ends are in the shape of isosceles triangles whose width is 5 meters and height is 2 meters. If water is being pumped in at a constant rate of 6 m^3/sec

.25 m/sec