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

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• Nov 19th 2012, 09:01 PM
theunforgiven
I don't even know where to start...
I've always enjoyed Algebra/Calculus. I wish I could say the same about problems. (Crying) Lately, I've been trying to improve my work problem solving skills, but I always find myself in need of someone's help, and that's why I'm here. I need help with this even problem. I don't know where or how to start. All I could think about was the Pythagorean theorem but that was about it. Some one please show me how to do this problem since there are more of it that I'm willing to try, but unless I havbe an example than I can get ideas from, I feel like I'm going nowhere. (Crying) Your help will be most appreciated.

http://img685.imageshack.us/img685/8...ehome1copy.jpg
• Nov 19th 2012, 09:49 PM
MarkFL
Re: I don't even know where to start...
Since we have a right triangle you did well to think of the Pythagorean theorem.

Now, if we label the horizontal leg of the triangle $x$, we know the vertical leg is fixed at 50 ft., and let's label the hypotenuse (the actual distance from the judge to the horse) as $h$. let all linear measures be in feet and time be in seconds. So we have:

$x^2+50^2=h^2$

Let's let $\theta$ represent the angle subtended by the vertical leg and the hypotenuse.

Now, we are also told that when $h=\frac{100}{\sqrt{3}}$ that $\theta=\frac{\pi}{6}$ and $\frac{d\theta}{dt}=-\frac{9}{10}$

Can you think of a way to relate $\theta$ to $h$ and the vertical leg?
• Nov 19th 2012, 10:01 PM
theunforgiven
Re: I don't even know where to start...
Take the derivative of the function above?
• Nov 19th 2012, 10:10 PM
MarkFL
Re: I don't even know where to start...
You need a relationship first...what function relates the angle subtended by the adjacent leg and hypotenuse of a right triangle?
• Nov 19th 2012, 10:19 PM
theunforgiven
Re: I don't even know where to start...
Quote:

Originally Posted by MarkFL2
You need a relationship first...what function relates the angle subtended by the adjacent leg and hypotenuse of a right triangle?

Ok let me think... is it Cos = Adjacent/hypotenuse? (Sorry, but I don't know how to format my equations the way you do yours.)
• Nov 19th 2012, 10:29 PM
MarkFL
Re: I don't even know where to start...
Yes, so we may then write:

$\cos(\theta)=\frac{50}{h}$

Now differentiate this with respect to time $t$, solve for $\frac{dh}{dt}$, then plug in the given values.

When you get spare time, do a search here and online for how to use $\LaTeX$, which will allow you to create nicely rendered mathematical expressions.(Wink)
• Nov 19th 2012, 10:40 PM
theunforgiven
Re: I don't even know where to start...
I know the derivative of cos (theta) is -sin(theta) (d theta/dt), but I'm sure how to differentiate h/50. I tried to use the quient rule but ended with some weird numbers..
• Nov 19th 2012, 10:46 PM
MarkFL
Re: I don't even know where to start...
You are correct about the left side. Try writing the right side as:

$50h^{-1}$

Now, just use the power and chain rules.
• Nov 19th 2012, 10:52 PM
theunforgiven
Re: I don't even know where to start...
Wow. Much easier! (Happy)

dh/dt = -30 feet per second? I don't know why I'm getting a negative value though...
• Nov 19th 2012, 11:01 PM
MarkFL
Re: I don't even know where to start...
Yes, we find:

$\frac{dh}{dt}=-30\,\frac{\text{ft}}{\text{s}}$

Now, to find $\frac{dx}{dt}$ you have two options:

i) relate the opposite and adjacent sides of the triangle.

ii) Use the Pythagorean theorem.

I recommend the first, as you may use given values, rather than computed values.
• Nov 19th 2012, 11:15 PM
theunforgiven
Re: I don't even know where to start...
So that's pretty much the end of part a! (Rock)

Ok give me five minutes and I'll see what I end up with!
• Nov 19th 2012, 11:26 PM
theunforgiven
Re: I don't even know where to start...
Quote:

Originally Posted by MarkFL2
i) relate the opposite and adjacent sides of the triangle.

I can't think of any relationship between opposite adjacent sides other than tan(theta)=opposite/adjacent. Am I on the right track?
• Nov 19th 2012, 11:32 PM
MarkFL
Re: I don't even know where to start...
Yes, exactly! Oh, and the negative result from part a) just means the distance between the judge and the horse is decreasing.
• Nov 19th 2012, 11:44 PM
theunforgiven
Re: I don't even know where to start...
Yeah I see.
• Nov 20th 2012, 12:05 AM
theunforgiven
Re: I don't even know where to start...
Does dx/dt equal -60? (Worried)
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