Functions of Several Variables, Area?

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CalleighMay

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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 guys ;)
 
hey there Calliegh:

Looks as if you are dealing with a total differential.

The area of the triangle can be found with \(\displaystyle A=\frac{1}{2}absin(C)\)

Differentiate w.r.t a, b, and C.

\(\displaystyle dA=\frac{1}{2}\left[(bsin(C))da+(asin(C))db+(abcos(C))dC\right]\)

Now, we know the given values, \(\displaystyle a=3, \;\ b=4, \;\ C=\frac{\pi}{4}, \;\ \;\ dC=\pm .02, \;\ da=db=\pm \frac{1}{16}\)

Plug them in and let me know what you get.
 
Thanks for the reply Galactus! ;)

Ok, so i plugged in a, b and c into the A equation and got:

A=1/2(3)(4)sin(pi/4)=4.7123

Then for the other equation i did:

in radian mode: (4sin(pi/4) = 2.8
in degrees mode: (3sin(4)=.20926
in radians mode: 3(4)cos(pi/4)=8.485

1/2(2.8+.20926+8.485)=5.75

Is that right? lol
 
No, I am sorry to say that is not right. Do not worry about the area formula in the beginning. Just look at the differentiated one.

Just plug the values into it.
 
k i did it again and got 6.71764, is this right now? lol It seems every time i do it i got a different answer
 
CalleighMay said:
k i did it again and got 6.71764, is this right now?
Click to expand...
You did which again? How? You got 6.71764 for what? (You have, I hope, switched your calculator to "radian" mode for the duration of this class, like they told you to back in your first calculus course...?)

The entire set-up and all the calculus and trig have been done for you. You've been given all of the values. You have one step remaining. Please show your work on this one last bit. Thank you! :D

Eliz.
 
k lol yeah i have it in radians mode ill show everything out.

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

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

An my calc in def in radians mode lol. What am i doing wrong AHHHH its driving me nuts!
 
CalleighMay said:
dA= 1/2((bsin(c) + asin(c) + abcos(c))
Click to expand...
How did you arrive at this formula? What were your steps and reasoning?

On the other hand, you could also try using the formula you were given earlier:

\(\displaystyle dA=\frac{1}{2}\left[(bsin(C))da+(asin(C))db+(abcos(C))dC\right]\)

\(\displaystyle a=3, \;\ b=4, \;\ C=\frac{\pi}{4}, \;\ \;\ dC=\pm .02, \;\ da=db=\pm \frac{1}{16}\)
Click to expand...
...and see where that leads.... :wink:

Eliz.
 
oh i need to plug the 1/16 in for dA and dB and .02 for dC after each term in the function?

Now i get .2398?
 
There ya' go. :D

Just make sure you express your answer as \(\displaystyle \pm 0.24\).
 
yay! so that's the answer? lol Are there units on that? Like percent maybe? Thank you SO much!
 
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