Math Help - 3 Bad Hard Problems

The period T of a clock pendulum (i.e., the time required for one back and forth movement) is given in terms of its lenght L by the equation:
T = [2PI][sqrt L/G], where g is the gravitational constant.

a) Assuming that the lenght of a clock pendulum can vary (say, due to temperature changes), find the rate of change of the period T with respect to the lenght L.

b) If L is in meters (m) and T is in seconds (s), what are the units of the rate of change in part (a)?

c) If a pendulum clock is running slow, should the lenght of the pendulum be increased or decreased to correct the problem?

Thanks in advance to anyone who help

2. Originally Posted by Collegegirl88
The period T of a clock pendulum (i.e., the time required for one back and forth movement) is given in terms of its lenght L by the equation:
T = [2PI][sqrt L/G], where g is the gravitational constant.
$T= 2\pi \sqrt{L/G}= 2\pi L^{1/2}G^{-1/2}$.

a) Assuming that the lenght of a clock pendulum can vary (say, due to temperature changes), find the rate of change of the period T with respect to the lenght L.
Find the derivative of T with respect to L, treating G as a constant. What is the derivative of $x^n$ for any real number, n?

b) If L is in meters (m) and T is in seconds (s), what are the units of the rate of change in part (a)?
The derivative is the limit of a fraction: T(L+h)-T(L)/h If T is in seconds and L and h are in meters, what units does that fraction have?

c) If a pendulum clock is running slow, should the lenght of the pendulum be increased or decreased to correct the problem?
Assuming that T and G are positive, is the derivative in (a) positive or negative? If it is positive, that means decreasing the length will decrease the period and so make the clock run faster. If it is negative, that means increasing the length will decrease the period and so make the clock run faster.