rate of change

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May 3rd 2006, 02:56 AM
dadon

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rate of change

hi all,

How about do I go to answer the following rate of change question.

A ladder of length 10metres is placed with one end of the horizontal ground and the other end against a vertical wall. Given the ladder makes an angle $\theta$ with the horizontal then it slips vertically down the wall at rate of 2 metres/sec. Find the rate at which the angle $\theta$ is decreasing at the distant when $\theta = \pi/4$ radian.

To help I have attached a picture.

I know the rate has to be minus (-2m/sec) as it is decreasing.

Can someone please explain the process step by step as I am totally lost

Thanks
May 3rd 2006, 05:34 AM
CaptainBlack

Quote:

Originally Posted by dadon

hi all,

How about do I go to answer the following rate of change question.

A ladder of length 10metres is placed with one end of the horizontal ground and the other end against a vertical wall. Given the ladder makes an angle $\theta$ with the horizontal then it slips vertically down the wall at rate of 2 metres/sec. Find the rate at which the angle $\theta$ is decreasing at the distant when $\theta = \pi/4$ radian.

To help I have attached a picture.

I know the rate has to be minus (-2m/sec) as it is decreasing.

Can someone please explain the process step by step as I am totally lost

Thanks

Let the top end of the ladder be $x$ m above the ground, then:

$<br /> \cos(\theta)=x/10<br />$ .

So differentiating this wrt $t$ gives:

$<br /> -\sin(\theta) \frac{d\theta}{dt}=\frac{1}{10}\ \frac{dx}{dt}<br />$ .

So now plug the values of $\theta$ and $\frac{dx}{dt}=-2$ in and solve for $\frac{d\theta}{dt}$ .

RonL
May 3rd 2006, 05:54 AM
Natasha1

CaptainBlack you are so clear in your explanations it's a pleasure to read your answers! really :-)
May 4th 2006, 01:14 PM
dadon

Thanks for that captain black!

Kind regards,

dadon