Question

Calculas 1 help
The minute hand of a certain clock is 4 in long. Starting from the moment when the hand is pointing straight up, how fast is the area of the sector that is swept out by the hand increasing at any instant during the next revolution of the hand?

Answer 1

use chain rule:

Answer 2

\[{d A \over d t}= {d A \over d \theta}\times{d \theta \over d t}\]

Answer 3

but where does all the info go? whats the equation

Answer 4

help please!

Answer 5

in 1 minute the the angle changes by 2pi/60 = pi/30
\[{d \theta \over d t}={\pi \over 30}\]

Answer 6

\[\ Area\ of\ sector = {\theta \over 2\pi}\pi r^2 = \frac \theta 2 r^2\]

Answer 7

so, \[{dA \over d \theta}={r^2 \over 2}\]

Answer 8

r is 4 inches.

So dA/d-theta = 16/2 = 8 sq. inches per radian

Answer 9

so,
dA/dt = 8*pi/30 = 4pi/15 square inches per minute ... which i believe is what you are looking for

Answer 10

yes that is what im looking for! so you just plug in what you want? im sorry its because my teacher does not know how to do this so I don't know why shes teaching the class

Answer 11

wait how did you get da/d0=r^2/2

Answer 12

from above, the formula for area of sector is
\[A(\theta)=\frac \theta 2 r^2\]

you can actually plug in r=4 right here:\[A(\theta) = 8 \cdot \theta\\\ \\\therefore {dA \over d \theta} =8\]