Calculus can be used to solve many problems. Probably the most important is solving real-life problems or situations. An example of this type of problem is a related rates problem. A related rates problem is a problem which involves at least two changing quantities and asks you to figure out the rate at which one is changing given sufficient information on all of the others. There are several key steps in setting up and solving related rates problems.

Some important and helpful things to do are as follows: 1. Draw a picture or make a model of the situation. 2. Label all quantities which can change as variables.

3. Identify in terms of the variables and their derivatives what is being asked and what is given. 4. Use the picture or model to write down a relation among the variables using an equation or something else which relates them. 5. Using the Chain Rule, implicitly differentiate both sides of the equation with respect to time t.

6. Substitute into the resulting equation all known values for the variables and their rates of change. 7. Solve for the required rate of change. 8.

Go back over your work and write up a solution. Example: One vehicle starts driving north and one vehicle starts driving east from an intersection. At the time the first vehicle is 5 miles north of the intersection and is traveling at 50 mph. At the same time, the second vehicle is 12 miles east of the intersection and is traveling at 65 mph. At that instant, what is the rate that the distance between the two vehicles is increasing Solution: The mathematical model of this situation is a right triangle. The legs of the triangle are increasing with time.

If we call the two legs x and y and the hypotenuse s, then all three quantities are changing. We are asked to find ds / dt at a particular time. (ds / dt ) is the rate at which the distance between the two car is increasing. We are given the values of x and y at this time as well as the values of dx / dt and dy / dt at this time. x = 5 miles (horizontal distance from intersection of car 1) y = 12 miles (vertical distance from intersection of car 2) dx / dt = 50 mph (speed of car 1) dy / dt = 65 mph (speed of car 2) Now we need to find an equation that relates the variables x, y, and s. By drawing a picture you can see that a right triangle is formed.

s y x So you can use the Pythagorean theorem as your relationship among the variables. Pythagorean Theorem: x 2 + y 2 = s 2. If we differentiate this with respect to t we obtain: From the equation you can see that we were given four of the six quantities (x, y, dx / dt , and dy / dt ) and we want to solve for ds / dt . There is one missing quantity s, which we can solve for by using the Pythagorean Theorem. x 2 + y 2 = s 2. (5) 2 + (12) 2 = s 2.

The result is that s = 13. To figure out what ds / dt equals you have to plug all the numbers back into the equation that was differentiated with respect to t and solve. Therefore: 2 (5) (50) + 2 (12) (65) = 2 (13) ds / dt . 2060 = 26 ds / dt ds / dt = 79. 23 mph. So at the instant in question, the distance between the vehicles is increasing at approximately 79.

23 miles per hour. This is just one type of problem that calculus is useful for solving. There are many other real life situations out there that depend on calculus to find the answer. Calculus is both a unique and definitely difficult way to solve problems.