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Let s(t) be a position function with velocity function v(t) = s'(t).
(a) If s(9) = 4 and , what is s(12)?
(b) If s(3) = -2 and , what is s(5)?
We know that
Plugging in the values we were given for s(9) and for the integral, we get
6 = s(12) – 4
so s(12) = 10.
A koala is climbing up and down its tree with velocity v(t) feet per minute, where positive values of v(t) indicate the koala is climbing up the tree.
At t = 0 minutes the koala is 5 feet above ground.
(a) If feet, how high is the koala after t = 4 minutes?
(b) If feet, how high is the koala at t = 7 minutes (use part (a))?
We're given that s(0) = 5 and , and asked to find s(4). Since
9 = s(4) – 5
and so s(4) = 14. This means the koala is 14 feet above ground after 4 minutes.
Since the koala is 14 feet above ground when t = 4, we know s(4) = 14.
We plug in these numbers:
At t = 7 minutes the koala is 11 feet above the ground.
A hummingbird flies away from its feeder with velocity v(t) feet per second and position s(t) feet away from its feeder.
(a) If s(4) = 5 feet and feet, find the hummingbird's distance from the feeder at time t = 0.
(b) If s(10) = 16 feet and feet, find s(5).
This problem is asking for s(0), so we plug in the numbers we're given and solve for s(0).
The hummingbird is 1 foot from the feeder when t = 0.
This is the same type of problem as (a), with different numbers.
The hummingbird is 18 feet from the feeder when t = 5.
(a) If and s(9) = -3, what is s(8)?
(b) If and s(7) = -4, what is s(3)?
(c) Given (a) and (b), what is ?
(d) Given (a) and (b), what is ?
(c) We know that
In (a) we were given that s(9) = -3, and in (b) we found that s(3) = -6.
(d) We know that
In (a) we found that s(8) = 1, and in (b) we were told that s(7) = -4. This means
(a) If s(2) = 7 and s(9) = 13, what is ?
(b) If s(2) = 13 and s(9) = 7, what is ?
(c) If s(2) = 13 and s(9) = 7 then what is ? (hint: one of the properties of integrals says how to change the limits of integration)
One of the properties of integrals lets us conclude that