Differentiating with respect to we have
Solving for , we get
Solution: The equation is , where is the
when . From (a),
Note: 1 kilometer = 1000 meters.
Solution: Let be the distance from the car to the
helicopter, and let be the distance from the car to the point
of the road directly below the helicopter. The problem states that
, and asks us to find the value of
at the moment when
. By the
at any time. Differentiating
and with respect to time, we get
b) Use the linearization to approximate .
c) Is your approximation greater than or less than the actual value? Justify your answer.
Solution: is negative for , so that the graph of is concave down on the domain of . This implies that the tangent line at the point (36,6) will lie above the graph of for . Since the graph of in (a) is this tangent line, our approximation, , will be greater than .
a) Find the intervals on which is increasing and those on which it is decreasing.
Solution: . The zeros of are . Since is differentiable for all , these are the only critical points of , so that we only need to test the sign of the derivative on the intervals , , , and . We have:
b) Find all the local maxima and local minima of .
Solution: We have critical points . Using the chart from (a) we find by the First Derivative Test that is a local minimum, is a local maximum, and is a local minimum.
c) Find the values of the global maximum and global minimum of over the interval .
Solution: We evaluate at the endpoints of the interval: and . Comparing these two values with the local minima and maxima we find that is the global minimum and is the global maximum on this interval.
d) Graph the function over the interval , indicating all the local and global extrema.
a) Express the acceleration in terms of the time variable and the constants , , and .
b) Assume now, that , (so , and ). Find the time intervals, where both the position and velocity are positive.
Solution: when for any integer . when for any integer . Thus both are positive when is in the intersection of these families of intervals, i.e., when , or for any intger .
c) Is the particle in part b) speeding up or slowing down, when both its position and velocity are positive? justify your answer.
Solution: When , which is clearly negative whenever is positive (they have opposite signs). So in particular, when and are both positive, will be negative. Thus velocity and acceleration will have opposite directions so that the particle will be slowing down.
a) Determine the intervals on which the function is concave upward and those on which it is concave downward.
Solution: exists for all , so that the only critical points of will be where , i.e., where , i.e., . Thus we have to test the sign of the second derivative on the intervals , , and . We have
b) Determine all points of inflection, if any. Justify your answer!
Solution: Since exists for all , our only candidates for inflection points are where , i.e., . From (a), we know that concavity changes at each of these points so that both are inflection points.