Functions of Several Variables, Area?

1. Aug 2, 2008

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.

2. Aug 3, 2008

futurebird

Forget about finding the maximum possible error for now and simply find the area of the triangle. How would you go about doing that? Describe the process.

3. Aug 3, 2008

CalleighMay

Well they give the triangle has one side 3 and one side 4, but doesn't give the opp side.

I thought maybe the angle between these two sides is 45 degrees (pi/4) but then how could i find the length of the other side? Wouldn't i just use 1/2b*h to find the area? To do that i need to be able to draw the triangle and find all angles and side lengths...

4. Aug 3, 2008

CalleighMay

I found some equation i think may be helpful but i think my answer's wrong since my professor laughed at me when he saw it...

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

dA= 1/2((4sin(pi/4) + 3sin(pi/4) + 3*4cos(pi/4)
dA= 1/2(2.83+2.12+8.49)
dA=1/2(13.44)
dA= 6.72

why is this wrong?

5. Aug 4, 2008

CalleighMay

Oops i forgot part of the equation, i tried it again and got + or - .24 does that sounds about right? Thanks ;)

6. Aug 5, 2008

7. Aug 5, 2008

CalleighMay

dA= 1/2((bsin(c) dA + asin(c) dB + abcos(c) dC)

dC=+/- .02
dA=dB=+/ -1/16

a=3
b=4
c=pi/4

dA= 1/2((4sin(pi/4)1/16 + 3sin(pi/4)1/16 + 3*4cos(pi/4).02)

=1/2(.3133)
=.156652 which when rounding is +/- .24

Is this not the correct way to do it?

8. Aug 5, 2008

HallsofIvy

Do NOT use small letters and capital letters interchangeably! In some formulas "A" and "a" might be used to mean different things. It is common in geometry to use small letters for lengths of sides of triangle and the corresponding capital letters for the angles opposite those sides. So your area formula is
A= (1/2)ab sin(C).

From that dA= (1/2)( b sin(C)da+ a sin(C)db+ ab cos(C)dC).

9. Aug 6, 2008

schroder

Hi! Well, you are using the correct formula, but not quite in the correct manner. Besides mixing metaphors, as HallsofIvy has mentioned, there are some other fundamental mistakes in your calculation method. What you are seeking to find is the total differential of the Area as a result of variations in several independent variables. This total differential will be expressed as a decimal ratio, or a percentage of the Area. So the variations in the independent variables must also be entered into the equation as decimal ratios, not in their individual units of measure. You cannot mix inches with radians and get a useful result, but you can mix the percentage of change in inches with the percentage of change in radians and get a percentage of change in Area. To that end, da = .0625/3 = .0208a, db = .0625/4 = .0156b and d Sin(C) = Sin .02/Sin (pi/4) = .028 Sin (C).
So now the formula becomes:

dA = (1/2)( b sin(C) .0208a + a sin(C) .0156b + ab .0208 sin(C) )

If you look closely at that equation you will see that it is now possible to factor out ab sin(C) from each term which leaves you with an elegant expression:

dA = (1/2) ab sin (C) (.0208 + .0156 + .0208)

dA = .0572 A so the Area can vary by 5.72% with the variations in the given parameters.

I hope this helps.