Related Rates and Derivatives... I need serious help

innanetaholic08

New member
Joined
Jan 1, 2007
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So, I got this Christmas Break packet and I've been putting it off because I have no clue how to do any of it and I didn't want to ruin my holiday, but now it's due tomorrow and I have yet to start. My grade sucks, so not turning it in would mean that I fail the class. I can't afford a failing grade on my report card. I am in sooo many clubs so I can't let my GPA fall. Anyone willing to help with some related rates and derivatives???? Please and Thank You.
 
Why did you put off doing your work if you are really worried about your grade?

In any case, we cannot help unless you post a problem. Not being mind readers we have no idea what your packet problems are.
 
sorry... Calculus really upsets me so I don't like to deal with it, so I pretend it doesn't exist.

Anyways my problem that I am attempting is:

Find the area od the largest rectangle that can be inscribed in a semicircle of radius 4 inches. I drew the semi circle with the rectangle in it. I have one side of the rectangle measuring 8-2x to account for the fact that there is x distance on either side of the rectangle on the flat side of the semicircle and the other side is y. I was just about to find the area of the semicircle. I think what I am doing is wrong though. Something is screaming at me that I'm not doing it right, but I can't think of any other way to do this.
 
Center your semicircle at the origin and inscribe your rectangle.

Draw a line from the origin up to the corner of the rectangle. You can

extend the y-axis upward and then break your rectangle into 2 halves.

This gives the area of the rectangle: A=2xy...[1]

The equation of the semicircle is \(\displaystyle \L\\y=\sqrt{16-x^{2}}\)..[2]

Sub [2] into [1]:

\(\displaystyle \L\\A(x)=2x(\sqrt{16-x^{2}})\)

Differentiate(think product rule), set to 0 and solve for x.

Your dimensions will follow.

circleyn1.gif


There's a small trick that may be helpful to check yourself.

The largest rectangle that can be inscribed in a semicircle of radius R has

an area of \(\displaystyle \frac{2}{\pi}\) that of the semicircle.

In other words, find the area of your semicircle, multiply by \(\displaystyle \frac{2}{\pi}\) and that'll give you the area of the largest rectangle that can be inscribed.

The area of your semicircle is \(\displaystyle \frac{\pi\cdot{4^{2}}}{2}=8{\pi}\)

\(\displaystyle 8{\pi}\cdot\frac{2}{\pi}=16\)

Go through the calculations and see if you get that.

Notice, that's also equal to the square of the radius.


It's a shame more people don't see mathematics for the beautiful thing it

is instead of having hangups about hating it or being afraid.

Relax, it'll be OK. And stop procrastinating.
 
wow thank you very very very very I could go on for hours but I still have a couple problems I need to finish. You are a real lifesaver and I am eternaly in your debt. I promise I will make it a rule not to procrastinate... well I wouldn't say promise, but I will seriously try. Again, thank you greatly.
 
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