Honors AP Calculus / Mr. Hansen	Name: _______________________________________
1/6/2007

Solution Key for OCQ (Oral Cumulative Quiz)

1.(a)	2.6 + ∫3^5 a(t) dt [This is a common AP question. Although some people had an equivalent answer last Friday, nobody boiled it down to this simple form.]
(b)	Avg. velocity = Ds/Dt = (s(7.5) – s(3))/(7.5 – 3), and by FTC1, numerator is ∫3^7.5 v(t) dt. Final answer: (∫3^7.5 v(t) dt)/4.5. Note: The numerator can be calculated by quadrature if we know a(t), since v(t) = 2.6 + ∫3^t a(t) dt. Direct substitution of v(t) into the expression ∫3^7.5 v(t) dt,        followed by double integration (integral inside another integral), would give        an answer. No initial condition for the displacement function s(t) is needed.
2.(a)	If a(t) is a function for which an explicit closed-form antiderivative is easily found, say for example a(t) = t – 2, I would request s(0) for simplicity. (Or, I might request s(1) or s(2), etc. It really does not matter.) However, if a(t) is a more difficult function, or a function defined only by means of a table of values, then I would request s for a value close to 4, perhaps s(3.95).
(b)	meters
(c)	If a(t) is a function for which an explicit closed-form antiderivative is easily found, one might be able to perform a double diffeq. solution. In other words, find a general solution for v(t) involving “+ C1,” and use the initial condition v(3) = 2.6 to find the particular solution for v(t). If the resulting v(t) also has an explicit closed-form antiderivative, then repeat the process. In other words, antidifferentiate v(t) to get s(t) as a general solution involving “+ C2,” and use the second initial condition, s(0) = whatever, to find a particular solution for s(t). Finally, plug in t = 4 to get an answer. However, it is often the case that the method above will not work. Perhaps a(t) or v(t) cannot be antidifferentiated in closed form. In this case, use Euler’s Method or, better yet, a one-step linear approximator, to estimate s(4) » s(3.95) + v(3.95)(.05). [Note: Points are deducted for using an “=” sign here.] If the teacher asks how we know what v(3.95) is, use a technique similar to #1(a) above and say v(3.95) = 2.6 + ∫3^3.95 a(t) dt.
3.(a)	definite integral, continuous, constant, independent variable (not a dummy variable)
(b)	antiderivative (by FTC2)
(c)	By FTC2, numerator simplifies to f (x) – f (4). Therefore, we are talking about the limit of a difference quotient, which simplifies to f ¢(4) by the definition of derivative. Note that we made use of the given fact that the limit exists. If the limit did not exist, the notation f ¢(4) would be meaningless. Final answer: f ¢(4).
(d)	As in part (c), the numerator simplifies to f (x) – f (4). Therefore, function R(x) represents a difference quotient between 4 and some other specified x value, namely the forward difference quotient with respect to 4. [It is OK to say “difference quotient,” which is what most of the students said last Friday, but it is better to be even more specific. Note that R(4) is undefined.] We all know (don’t we?) that “difference quotient” means the same as “average rate of change” or “secant slope.” Since f follows a straight line in an interval that covers [4, 4.001], the instantaneous rate of change at 4.001, given to be 2, must equal the difference quotient from 4 to 4.001, which was given to be R(4.001). Final answer: R(4.001) = 2.
4.(a)	P1 is a curve (a set of ordered pairs), but the question asked about P1(t), which is an ordered pair.        If you prefer, you could say “a vector with initial point (0, 0) and terminal point (x1(t), y1(t)).”        However, the concept is easier to visualize if you think of P1(t) as an ordered pair.
(b)	The key insight is that P1 and P2 are inverse relations with respect to x and y. We discussed in class how the instantaneous rates of change for a relation and its inverse are reciprocals—not at the same x value, of course, but at values that correspond symmetrically across the line y = x, and those corresponding points of P1 and P2 were given to have the same t value. The ordered pairs for P1(t) and P2(t) are always closely related: if one is (a, b), the other is (b, a). We could use the parametric chain rule to find the tangent slope at various points on curve P1, but that is not really necessary in this problem. It suffices to remember that dy/dx at a point of P1 equals dx/dy (i.e., the reciprocal) at a corresponding point of P2. To summarize: The tangential slope at a point on P1 equals the reciprocal of the tangential slope at a mirror-image point on P2. (As for situations involving undefined slope or slope of 0, we note that if one tangent is horizontal, the other is vertical.) However, the problem asked us to compare a P1 tangent with a P2 normal. Since the P2 normal has a slope that is the opposite reciprocal of a P2 tangent slope, we have the opposite reciprocal of a reciprocal. Final answer: The P1 tangent slope equals the opposite (i.e., additive inverse) of the P2 normal slope. If the P1 tangent is vertical, then the P2 normal is also vertical.