Problem of the Week#7: 2001-2002

Problem Seven: 2001-2002

Mean Value Theorem: Calculus I (AB)

If the function satisfies the Mean Value Theorem at x = 2 on the closed interval [1, a], find a.

Difficulty:

Solution:

If the Mean Value Theorem is satisfied at x = 2, that means that the slope of the tangent line there (the instantaneous rate of change of f) is equal to the slope of the secant line on the interval [1, a] (the average rate of change of f over that interval). Therefore, f '(2) is equal to the slope of the secant line (found by the change in y divided by the change in x--the same slope formula you have used since you were an embryo).

Which value for a is correct? It can't be 1, or our interval would be [1,1], which is no interval at all. The correct answer is 3. If you go back and calculate the secant slope for f on [1,3], you get 11, which is f '(2).

Implicit Differentiation: Calculus II (BC)

Find the equations of all the tangent lines to the graph of when x = 0.

Difficulty:

Solution:

First order of business: what points are on the graph when x = 0? Find out by plugging in x = 0.

To solove this equation, use the quadratic formula or your graphing calculator to get y = .76393202 and 5.236068. You shouldn't round off the decimals until the end of the problem, so use all of those babies in your calculations. Now, you know that the points (0, .76393202) and (0, 5.236068) are both y-intercepts for this equation, which is clearly not a function, as this example alone illustrates a failure of the vertical line test. Now, we have the points at which the tangent lines will be tangent. The only other thing we need to create a line besides a point is the slope of that line. Because we are finding tangent lines, the slopes will be given by the derivative. Because this is not a function of one variable (you can't solve it for either x or y), you have to use implicit differentiation. Don't forget you have to use the product rule when differentiating 3xy.

Now, plug in the (x, y) pair from above to get the slopes at those points. Slope of the tangent line for point (0, .76393202): -.605572809. Slope at point (0, 5.236068): -2.394427191. Now, use point-slope form to write both of those lines for your final answer. The lines do not have to be in standard form. The problem does not require it, so don't waste the time.

Superbowl of Calculus