# Math Help - I don't even know where to start...

1. ## Re: I don't even know where to start...

I would write:

$\tan(\theta)=\frac{x}{50}$

Differentiating with respect time time $t$, we find:

$\sec^2(\theta)\frac{d\theta}{dt}=\frac{1}{50}\cdot \frac{dx}{dt}$

$\frac{dx}{dt}=50\sec^2(\theta)\frac{d\theta}{dt}$

Now plug in the given values:

$\frac{dx}{dt}=50\sec^2\left(\frac{\pi}{6} \right)\left(-\frac{9}{10} \right)$

So what do we get?

2. ## Re: I don't even know where to start...

I don't know how to plug in secant (squared) into my calculator... :S

EDIT: -59.99?

3. ## Re: I don't even know where to start...

You shouldn't need to use you calculator. The given angle is a "special angle" for which we should know all of the trig functions at this angle.

What is $\cos\left(\frac{\pi}{6} \right)$ ?

If you know this, then since $\sec(\theta)\equiv\frac{1}{\cos(\theta)}$ you can invert, then square to get the value you want.

4. ## Re: I don't even know where to start...

Originally Posted by MarkFL2
You shouldn't need to use you calculator. The given angle is a "special angle" for which we should know all of the trig functions at this angle.

What is $\cos\left(\frac{\pi}{6} \right)$ ?

If you know this, then since $\sec(\theta)\equiv\frac{1}{\cos(\theta)}$ you can invert, then square to get the value you want.
And that's exactly what I did. I ended up with -59.99 (or -60). Oh and $\cos\left(\frac{\pi}{6} \right)$ = square root of 3 over 2.

5. ## Re: I don't even know where to start...

Yes, so what do you get when you invert and square?

6. ## Re: I don't even know where to start...

Originally Posted by markfl2
yes, so what do you get when you invert and square?
-60. Hope that's what you are asking for.

7. ## Re: I don't even know where to start...

That is the correct final answer, but I was asking what you got when you invert and square the value you found for the cosine function.

8. ## Re: I don't even know where to start...

Oh my bad. I got 0.66666666667

9. ## Re: I don't even know where to start...

What I mean is:

invert: $\frac{2}{\sqrt{3}}$

square: $\left(\frac{2}{\sqrt{3}} \right)^2=\frac{4}{3}$

Thus, we know:

$\sec^2\left(\frac{\pi}{6} \right)=\frac{4}{3}$

10. ## Re: I don't even know where to start...

Stupid me! But thank you, sir, I truly appreciate all your help. You are a true math master! *bows down*