1. ## [SOLVED] Applications of trig

A rotating light L is situated at sea 180 metres from the point P on a straight shoreline. The light rotates through one revolution every 10 seconds. Show that the rate at which a ray of light moves along the shore at a point 300 metres from P is 136pi m/s.

Thanks!

2. Related rates nearly always depend on the chain rule, so you might want to try filling up this pattern...

... where straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case time), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

Assuming the shore is perpendicular to the line PL, we can use basic trigonometry to express x, the distance of the beam along the shore from L, in terms of theta the angle of the beam away from the line PL...

So differentiate with respect to the inner function, and the inner function with respect to t...

Spoiler:

Use a right-triangle diagram to see that sec(arctan(t)) = $\displaystyle \sqrt{1 + t^2}$

_________________________________________
Don't integrate - balloontegrate!

Balloon Calculus: Standard Integrals, Derivatives and Methods

Balloon Calculus Drawing with LaTeX and Asymptote!

3. Originally Posted by ipokeyou
A rotating light L is situated at sea 180 metres from the point P on a straight shoreline. The light rotates through one revolution every 10 seconds. Show that the rate at which a ray of light moves along the shore at a point 300 metres from P is 136pi m/s.

Thanks!
given ...

$\displaystyle \frac{d\theta}{dt} = \frac{2\pi}{10}$ rad/s

$\displaystyle x = 300$ m

relationship between $\displaystyle x$ and $\displaystyle \theta$ ...

$\displaystyle \theta = \arctan\left(\frac{x}{180}\right)$

take the time derivative of the above equation, sub in your given values and solve for $\displaystyle \frac{dx}{dt}$

4. Originally Posted by skeeter
given ...

$\displaystyle \frac{d\theta}{dt} = \frac{2\pi}{10}$ rad/s

$\displaystyle x = 300$ m

relationship between $\displaystyle x$ and $\displaystyle \theta$ ...

$\displaystyle \theta = \arctan\left(\frac{x}{180}\right)$

take the time derivative of the above equation, sub in your given values and solve for $\displaystyle \frac{dx}{dt}$
how do i take the time derivative? because you cant differentiate
$\displaystyle \theta = \arctan\left(\frac{x}{180}\right)$
with respect to t?

and also if $\displaystyle x = 300$ then cant we just sub this into the above equation?

sorry, i dont think i'm on the right track here..

5. Originally Posted by ipokeyou
how do i take the time derivative? because you cant differentiate
$\displaystyle \theta = \arctan\left(\frac{x}{180}\right)$
with respect to t?

and also if $\displaystyle x = 300$ then cant we just sub this into the above equation?

sorry, i dont think i'm on the right track here..
Post #2 said to use the chain rule and then gave you all you needed. Have you been given examples or done similar questions so that you understand what this means?

From the chain rule: $\displaystyle \frac{dx}{dt} = \frac{dx}{d \theta} \cdot \frac{d \theta}{dt}$.

6. Originally Posted by mr fantastic
Post #2 said to use the chain rule and then gave you all you needed. Have you been given examples or done similar questions so that you understand what this means?

From the chain rule: $\displaystyle \frac{dx}{dt} = \frac{dx}{d \theta} \cdot \frac{d \theta}{dt}$.
yes i have done similar questions. but i dont really understand what the diagram means...

7. Originally Posted by ipokeyou
how do i take the time derivative? because you cant differentiate
$\displaystyle \theta = \arctan\left(\frac{x}{180}\right)$
with respect to t?
You can, using the chain rule - since x and theta are both functions of t.

Originally Posted by ipokeyou
and also if $\displaystyle x = 300$ then cant we just sub this into the above equation?
Yes, to find theta for x = 300. Sub into the differentiated equation to find dx/dt for x = 300 and d theta / dt = 2 pi / 10.

Originally Posted by ipokeyou
yes i have done similar questions. but i dont really understand what the diagram means...
Use whatever you can...

8. Originally Posted by ipokeyou
yes i have done similar questions. but i dont really understand what the diagram means...
If you understand problems involving related rates, then you ought to understand how the chain rule is used. So substitute the things told to you in this thread into the 'formula' I gave in my previous post.

9. Originally Posted by ipokeyou
how do i take the time derivative? because you cant differentiate
$\displaystyle \theta = \arctan\left(\frac{x}{180}\right)$
with respect to t?
sure you can ...

$\displaystyle \frac{d}{dt}\left[\theta = \arctan\left(\frac{x}{180}\right)\right]$

$\displaystyle \frac{d\theta}{dt} = \frac{\frac{1}{180}}{1 + \left(\frac{x}{180}\right)^2} \cdot \frac{dx}{dt}$

10. Originally Posted by skeeter
sure you can ...

$\displaystyle \frac{d}{dt}\left[\theta = \arctan\left(\frac{x}{180}\right)\right]$

$\displaystyle \frac{d\theta}{dt} = \frac{\frac{1}{180}}{1 + \left(\frac{x}{180}\right)^2} \cdot \frac{dx}{dt}$
oh. that's what i did. but i thought it was wrong... sorry about that. but thanks!