I don't even know where to start...

**MarkFL** · November 20th 2012, 12:12 AM

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?

**theunforgiven** · November 20th 2012, 12:16 AM

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

EDIT: -59.99?

**MarkFL** · November 20th 2012, 12:25 AM

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.

**theunforgiven** · November 20th 2012, 12:26 AM

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.

**MarkFL** · November 20th 2012, 12:30 AM

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

**theunforgiven** · November 20th 2012, 12:32 AM

Originally Posted by markfl2

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

-60. Hope that's what you are asking for.

**MarkFL** · November 20th 2012, 12:40 AM

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.

**theunforgiven** · November 20th 2012, 12:46 AM

Oh my bad. I got 0.66666666667

**MarkFL** · November 20th 2012, 12:51 AM

What I mean is:

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

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

Thus, we know:

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

**theunforgiven** · November 20th 2012, 12:57 AM

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

**MarkFL** · November 20th 2012, 01:00 AM

Glad to be of assistance!

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