not, and recall that v(t) = x′(t) and a(t) = x′′(t). First, take the derivative of x(t), set it equal to
zero, and solve for t: x′(t) = 9t^2 − 4 = 0 and t = . To determine whether the particle changes
at that time, take the second derivative and determine the value of the second derivative at t =
: x(t) = 18t and x = 12. Since, x′′(t) is not zero when x′(t) is, the particle is changing
direction at t = .
- B A 20-foot ladder slides down a wall at 5 ft/sec. At what speed is the bottom sliding out when
the top is 10 feet from the floor (in ft/sec)?
First, let’s make a sketch of the situation.
We are given that = −5 (it’s negative because the ladder is sliding down and it’s customary
to make the upward direction positive), and we want to find when y = 10.
We can find a relationship between x and y using the Pythagorean Theorem. We get x^2 + y^2 =
400.
Now, taking the derivative with respect to t, we get
2 x + 2y = 0, which can be simplified to x = −y
Next, we need to find x when y = 10.
Using the Pythagorean Theorem,
x^2 + 10^2 = 400, so x = ≈ 17.321
