The third midterm is here.
Problem 1. Consider the region bounded below by the curve and above by the -axis. A rectangle is inscribed in this region in such a way that one of its sides lies on the -axis and the other two corners are on the parabola. Find the dimensions of the sides of the rectangle that maximize its area.
Suppose that and are the dimensions of the sides of the rectangle, with the length of the horizontal sides. Then the area of the rectangle is , and in order to apply calculus to maximize this function, we first express and in terms of .
We have that and . Hence, if denotes the area of the rectangle, as a function of , we have that
Note that, by our convention, . We have , so , or , or
Since , it must be the case that . Since , it follows that this value of gives us a local maximum of the function , and therefore the only maximum in the interval , which is the range of values that can take and be a length of a rectangle as required.
It follows that
and the maximum area is .
Problem 2. Two (very long) roads cross each other at a point . One runs north to south, and the other runs east to west. Car is driving north along the first road, and car is driving east along the second road. At a particular time, car is 30 miles to the north of and traveling at 50 m/h, while car is 40 miles to the east of and traveling at 65 m/h. How fast is the distance between the two cars changing?
The graph above (click to enlarge) shows the situation described in the problem. Given an arbitrary time , let’s denote by the distance from to the point of intersection of the roads, by the distance from to , and by the distance between and , all computed at time .
The problem is asking us to compute at the time that makes and , where units are measured in miles, and time is measured in hours.
Note first that , by the Pythagorean theorem. Then . or
At the time that we are interested in, we have , , , and , so
Problem 3. Let . Find a point between and with the property that coincides with the slope between the endpoints of on .
The slope of the line between the endpoints of on is simply
so we need to solve the equation
For this, note that , so and we have
, or , or .
In the graph above (click to enlarge) I have displayed in different colors the line and the line tangent to at . I have also shown the line and the curve . Note how the tangent line is indeed parallel to the given secant line.
Problem 4. The equation obviously has solution . Show the values of obtained by applying Newton’s method to this equation, starting with .
Newton’s method begins with a guess (in this case ) and iteratively improves it, approaching a solution of the equation In this case, . The iterative process is given by the equations
In our case, , so and
so , i.e., in our iterative process we have
This means that:
- , and
Problem 5. Find all functions such that
From results in class, we know that if for all in the common domain of and then (on each interval where both and are defined) there is a constant such that .
What this means for the problem at task is that it suffices to guess a function with the given derivative, because then any other function with the same derivative will simply differ from our guess by a constant. From the derivatives studied through the course, we easily see now that
To find , we use that : From the formula above, we have that
or , so
The two graphs below (click to enlarge) show first and then , with a little circle indicating the position of the point on the graph.