# rate of change

• Jun 4th 2011, 06:26 AM
Punch
rate of change
http://i952.photobucket.com/albums/a...g/P5250091.jpg

The diagram shows an isosceles triangle $ABC$ with fixed length $AB$ and $AC$ of $10cm$ each and Angle $ACB =$Angle $ABC = \theta$radians.

$A$ is a variable point which is at a height $h cm$ directly above the point $O$ while $B$ and $C$ are variable points which move horizontally along the line $l$.

Given that $A$ descends vertically towards the point $O$ such that the area of triangle $ABC$ is decreasing at a constant rate of $0.7cm^2/s$, determine, at the instant when $A$ is $6cm$ above $O$, the rate of change of $\theta$

$A=\frac{1}{2}(20cos\theta)h$

$=10hcos\theta$

$\frac{dA}{d\theta}=-10hsin\theta$

$when h=6, sin\theta =\frac{6}{10}$

$\theta=0.6435 rad$

$\frac{dA}{d\theta}=-10(6)(sin0.6435)$
$=-36$

$\frac{d\theta}{dt}=(\frac{dA}{dt})(\frac{d\theta}{ dA})$

$=(-0.7)(\frac{1}{-36})$

$=\frac{7}{360}$

And another part of the question asks : Find the rate at which C is moving away from O. I could not start
• Jun 4th 2011, 07:20 AM
skeeter
note that the area of a triangle can be calculated from the side-angle-side formula

$A = \frac{1}{2}ab\sin{\phi}$

where a, b are the adjacent sides and $\phi$ is the angle between them

$A = \frac{1}{2} \cdot 10^2 \sin(\pi - 2\theta)$

using the sine difference identity ...

$A = 50 \sin(2\theta)$

now take the time derivative, sub in the given/calculated values and determine the value of $\frac{d\theta}{dt}$
• Jun 4th 2011, 07:26 AM
Punch
Quote:

Originally Posted by skeeter
note that the area of a triangle can be calculated from the side-angle-side formula

$A = \frac{1}{2}ab\sin{\phi}$

where a, b are the adjacent sides and $\phi$ is the angle between them

$A = \frac{1}{2} \cdot 10^2 \sin(\pi - 2\theta)$

using the sine difference identity ...

$A = 50 \sin(2\theta)$

now take the time derivative, sub in the given/calculated values and determine the value of $\frac{d\theta}{dt}$

but i cant seem to spot the mistake in my workings
• Jun 4th 2011, 07:47 AM
skeeter
Quote:

Originally Posted by Punch
but i cant seem to spot the mistake in my workings

note that h is changing. you should have dh/dt in your derivative also ... problem is, that value was not given.
• Jun 4th 2011, 08:40 AM
Punch
Quote:

Originally Posted by skeeter
note that h is changing. you should have dh/dt in your derivative also ... problem is, that value was not given.

yes, i should have expressed h in terms of $\theta$, thanks
• Jun 4th 2011, 06:29 PM
Punch
hello, the question has a second part which i couldn't start, could i get some help with it? i think its linked to the first part.

The other part of the question asks : Find the rate at which C is moving away from O. I could not start
• Jun 4th 2011, 10:18 PM
Corpsecreate
Did you get the first part done alright?

The second part requires you to look at the length OC. Call the length OC a new variable (I called it x) and try to solve for $dx/dt$