Here are some word problems that involve antidifferentiation…
A rock is dropped from the top of a 400 foot cliff.
It’s velocity at time t seconds is:
v(t) = -32t feet per second.
A) Find s(t), the height of the rock above the ground at time t
s(t) = S v(t) dt
In English, this means that s(t) is the antiderivative of v(t).
It is. Trust me.
So, in order to find s(t) we must take the antiderivative of our v(t)
s(t) = S -32t dt
= -32/2 * t^2 + C
= -16t^2 + C
But we are not finished yet.
Because we know that the rock was dropped off a 400 foot cliff, this means that when we started, the rock was at a position of 400.
In other words, when t = 0, s(t) = 400.
Therefore, we can put 0 in for t and 400 in for s(t) and solve for C.
s(t) = -16t + C
400 = -16(0) + C
400 = C
This makes our final answer:
s(t) = -16t + 400
B) How long will it take the rock to reach the ground?
When the rock hits the ground, it is 0 feet off the ground.
Therefore, s(t) = 0. So we set s(t) equal to zero and solve for t.
s(t) = 0
-16t +400 = 0
-16t = -400
t = 25 seconds
C) What will be its velocity when it hits the ground?
We already found the time when it hits the ground in part b
In order to find the velocity, all we need to do is plug the t (=25) that we found into the equation for velocity.
v(t) = -32t
v(25) = -32(25)
= 800 feet per second.
After t hours of opperation a coal mine is producing coal at the rate of 40 + 2t -1/5*t^2
Find a formula for the total output of the coal mine after t hours of operation
Recall, whenever you would see the word “rate” in a problem, you should automatically think “derivative”
Well, the situation is similar here only this time they give us the rate aka, they give us the derivative
So in order to find the formula for total output, all we need to do is take the antiderivative of the formula that they gave us.
S40 + 2t -1/5*t^2 dx
= 40t + t^2 – 1/3 * 1/5 t^3 + C
= 40t + t^2 – 1/15 * t^3 + C