Calculus Tutorial
Trajectory
Intro
One of the most useful applications of Calculus in everyday life is finding the rate of change of a function. Finding the Trajectory of a moving object is a direct example of such. This tutorial will help explain how to use basic calculus methods to solve an otherwise overwhelming problem.
Sample Problem
A circus clown is shot out of a cannon to a landing platform across the stage. The distance from the clown to the ground is given by the formula 
where 
is the number of seconds the clown is in the air from the moment it shot, and 
is the number of feet above the ground the clown is at seconds.
- Find a formula for the velocity function.
- At what speed did the clown emerge from the cannon?
- How high was the clown above the ground the moment he was fired?
- The clown will go up and then down. How many seconds has he been flying when he reaches the highest point of his trajectory?
- How high does the clown go?
- How long has the clown been in the air when he lands?
Solution
- Since gives us our position at time , the change in position would give us our velocity or speed. The change in a function is given by its derivative so to find the velocity function we simply take the derivative of the position function. Using the power rule for derivatives we know that
- Since we have the velocity function now, we can simply plug in a
for time to get the speed at the moment of firing and get 
feet/sec. - Likewise, since we know the position function, we can plug in a for time and get an initial position of
feet. - Since he is moving against gravity, there is a point at which he achieves maximum height at a speed of . Knowing this, we can set the velocity function equal to and solve for to find the time at which this occurs.
Using some basic algebra we find that 
so 
seconds is our answer. - Now that we have figured out the time at which the clown achieves a max peak, we can plug the time into the position function to learn the height.
feet. - When the clown lands the position will be . Setting the position function to gives us
. Since this is a quadratic equation we need to either factor this or use the quadratic formula to find our values. This particular equation does not factor so we will use the quadratic formula 
. Plugging in 
for 
, 
for 
, and 
for 
we get 
which can be represented as approximately 
and 
. We can’t have a negative time so seconds is our answer.
About The Author
| Math Expert |
| My name is Josh Clark. I have worked at my college, Mars Hill University in various math related positions. I have worked as a personal tutor through student services helping fellow students who may have documented disabilities. One girl I helped who had dyslexia and sever test anxiety had gotten th... |
7 Subjects
KnowRo Tutor
1 Tutorial
Suggested Tutors for Calculus Help
