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

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• Nov 20th 2012, 12:12 AM
MarkFL
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?
• Nov 20th 2012, 12:16 AM
theunforgiven
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? (Worried)
• Nov 20th 2012, 12:25 AM
MarkFL
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.
• Nov 20th 2012, 12:26 AM
theunforgiven
Re: I don't even know where to start...
Quote:

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. (Itwasntme)
• Nov 20th 2012, 12:30 AM
MarkFL
Re: I don't even know where to start...
Yes, so what do you get when you invert and square?
• Nov 20th 2012, 12:32 AM
theunforgiven
Re: I don't even know where to start...
Quote:

Originally Posted by markfl2
yes, so what do you get when you invert and square?

-60. Hope that's what you are asking for. (Worried)
• Nov 20th 2012, 12:40 AM
MarkFL
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.
• Nov 20th 2012, 12:46 AM
theunforgiven
Re: I don't even know where to start...
Oh my bad. I got 0.66666666667 (Giggle)
• Nov 20th 2012, 12:51 AM
MarkFL
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:

$\sec^2\left(\frac{\pi}{6} \right)=\frac{4}{3}$
• Nov 20th 2012, 12:57 AM
theunforgiven
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*
• Nov 20th 2012, 01:00 AM
MarkFL
Re: I don't even know where to start...