Functions of Several Variables, Area?

[CalleighMay]

Hello, my name is Calleigh and i am new to the forum! I am in Calculus II and have a few questions on some problems. I am using the textbook Calculus 8th edition by Larson, Hostetler and Edwards. Could someone please help me?

 

The problem is on pg 922 in chapter 13.5 in the text, number 32. It reads:

 

A triangle is measured and two adjacent sides are found to be 3 inches and 4 inches long, with an included angle of pi/4. The possible errors in measurement are 1/16 inch for the sides and .02 radian for the angle. Approximate the maximum possible error in the computation of the area.

 

I haven't had any problems like this in class, so i don't know what to do. My professor suggested drawing a picture, but i haven't the slightest clue even where to begin. My professor explained it to me but i didn't understand it at all... any help would be greatly appreciated. Thanks guyssss

[booker]

Normally you'd post this in the homework folder, and you'd be required to give it a good attempt. The more work you put in, the more you get in help get, seems to be the general idea.

 

First, what's the formula for the area of the triangle when given two sides, and their interior angle?

[CalleighMay]

my friends and i have come up with the following but not sure if we're right,

 

dA= 1/2((bsin© dA + asin© dA + abcos© dA)

 

When i did this, a=3, b=4, c=pi/4

 

i got + or - .24

 

Does this sound about right? What would units be, percent? thanks

[booker]

That's an interesting attempt. Is Calculus II about multivariable equations? I thought that was III. If not, that's what this problem is leading into.

 

Your solution is about what changes happen in A when you change one variable and hold the rest constant. Then do the others, in turn, and add up the results. It's called the total derivative.

 

In terms of differentials, which is a first approximation of the total error, in this case

 

[math]dA = \frac{\partial A}{\partial a}da + \frac{\partial A}{\partial b}db + \frac{\partial A}{\partial \theta}d\theta[/math]

 

Is this notation familiar to you?

 

Maybe you're making too big a deal of this thing when the maximum calculated area is simply

 

[math] A_{max} = (1/2)(3+0.625)(4+0.625)sin((\pi/2)+0.02)[/math]

 

and the minumum

 

[math] A_{min} = (1/2)(3-0.625)(4-0.625)sin((\pi/2)-0.02)[/math]

 

So the maximum error in A would be the largest of Amax-A or A-Amin.

Edited August 5, 2008 by booker

[CalleighMay]

so does that come out to .24 as well? Is that correct? lol

[booker]

These two equations,

 

 

[math] A_{max} = (1/2)(3+0.625)(4+0.625)sin((\pi/2)+0.02)[/math]

 

[math] A_{min} = (1/2)(3-0.625)(4-0.625)sin((\pi/2)-0.02)[/math]

 

 

should read

 

[math] A_{max} = (1/2)(3+0.0625)(4+0.0625)sin((\pi/4)+0.02)[/math]

 

[math] A_{min} = (1/2)(3-0.0625)(4-0.0625)sin((\pi/4)-0.02)[/math]