Author Topic: Applications of differentiation question (Read 634 times) Tweet Share

cns1511 · « **on:** May 05, 2009, 04:39:22 pm »

Help please.

/0 · « **Reply #1 on:** May 05, 2009, 04:54:02 pm »

a) The minute hand of a clock sweeps out an angle of $2\pi$ every 1 hour at a constant rate, so $\frac{d\theta}{dt} = \frac{2\pi}{1} = 2\pi\ rad/h$

b) We want $\frac{ds}{dt}$

We can find it by using the chain rule: $\frac{ds}{dt} = \frac{ds}{d\theta} \times \frac{d\theta}{dt}$

$s=r\theta \Rightarrow \frac{ds}{d\theta} = r$

$\therefore \frac{ds}{dt} = (r) \times 2\pi = 2\pi r = 2\pi (10) = 20\pi \ cm/h$

c) We want $\frac{dA}{dt}$ so we can use $\frac{dA}{dt} = \frac{dA}{d\theta} \times \frac{d\theta}{dt}$

$A = \frac{1}{2}r^2\theta \Rightarrow \frac{dA}{d\theta} = \frac{1}{2}r^2$

$\therefore \frac{dA}{dt} = \left(\frac{1}{2}r^2\right)(2\pi) = \pi r^2= \pi (10)^2= 100\pi\ cm^2/h$

Actually! There's a much easier way to do b) and c) using circle formulas.
The minute hand covers the entire circumference of $2\pi r$ in 1 hour, and it covers an area of $\pi r^2$ in 1 hour, so there are your rates.

cns1511 · « **Reply #2 on:** May 05, 2009, 04:57:30 pm »

Thanks

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