MATH 180 FALL 1995 10:00 SECTION FONG

ASSIGNMENT #22

4.2, #32. f(x) = 2x^{3} − 2x^{2} + 1 has derivative f′(x) = 6x^{2} − 4x. In particular, f′(1) = 6 − 4 = 2. Thus the tangent line at (1, 1) has slope 2. This is the line y − 1 = 2(x − 1) or y = 2x − 1.

4.2, #38. The graph of the parabola f(x) = x^{2} − 2x + 4 suggests there are two lines through the origin which are tangent to the graph. One way to find these lines: (i) Write the equation of the tangent line at the point (a, f(a)). (ii) Determine if the tangent line goes through (0, 0). Indeed, f′(x) = 2x − 2, so that f′(a) = 2a − 2. Thus the tangent line at (a, f(a)) = (a, a^{2} − 2a + 4) has slope 2a − 2. The equation of this line is

y − (a^{2} − 2a + 4) = (2a − 2)(x − a).

The origin (0, 0) is on the line if −(a^{2} −2a+4) = (2a−2)(−a). This reduces to a^{2} −4 = 0, or a = ±2. If a = 2, the point is (2, 4), the slope is 2, and the tangent line is y = 2x. If a = −2, the point is (−2, 12), the slope is −6, and the tangent line is y = −6x.

4.2, #42. T = 2π\sqrt{\frac{ℓ}{g}} = \frac{2π}{\sqrt{g}} ℓ^{−1/2} and \frac{dT}{dℓ} = \frac{π}{\sqrt{g}} ℓ^{−1/2}. Thus dT/dℓ is positive, which

implies the period increases as the length of the pendulum increases.

4.3, #1, 15.

\begin{aligned}y & = 5t^{2} + 4e^{t}, \\ y′ & = 10t + 4e^{t}, \end{aligned}  \begin{aligned}f(x) & = x^{3} + 3^{x} \\ f′(x) & = 3x^{2} + (ln 3)3^{x} \end{aligned}