# rate of change

• May 3rd 2006, 02:56 AM
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
• May 3rd 2006, 05:34 AM
CaptainBlack
Quote:

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

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
• May 3rd 2006, 05:54 AM
Natasha1