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 $\displaystyle \theta $ with the horizontal then it slips vertically down the wall at rate of 2 metres/sec. Find the rate at which the angle $\displaystyle \theta $ is decreasing at the distant when $\displaystyle \theta = \pi/4$ 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

Quote:

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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 $\displaystyle \theta $ with the horizontal then it slips vertically down the wall at rate of 2 metres/sec. Find the rate at which the angle $\displaystyle \theta $ is decreasing at the distant when $\displaystyle \theta = \pi/4$ 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

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

$\displaystyle
\cos(\theta)=x/10
$.

So differentiating this wrt $\displaystyle t$ gives:

$\displaystyle
-\sin(\theta) \frac{d\theta}{dt}=\frac{1}{10}\ \frac{dx}{dt}
$.

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

RonL

CaptainBlack you are so clear in your explanations it's a pleasure to read your answers! really :-)

Thanks for that captain black!

Kind regards,

dadon