Using Partial Derivatives to estimate error

[Nikstykal]

Homework Statement

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The area of a triangle is (1/2)absin(c) where a and b are the lengths of the two sides of the triangle and c is the angle between. In surveying some land, a, b, and c are measured to be 150ft, 200ft, and 60 degrees. By how much could your area calculation be in error if your values for a and b are off by half a foot each and your measurement of c off by 2 degrees? See figure below

Homework Equations

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dA(x,y,z) = dA/dx dx + dA/dy dy + dA/dz dz
A = 1/2*absin(c)
a = 150+-0.5 ft, b = 200+-0.5 ft, c = 60+-2 degrees

The Attempt at a Solution

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I first used the chain rule on the Area to get dA

dA = 0.5bsin(c)*da + 0.5asin(c)*db + 0.5abcos(c)*dc

Setting da = 0.5, db = 0.5, and dc = 2 and plugging in my values I get a completely wrong answer (the actual answer is 338 ft^2). Something felt off about the dc, because it varied so differently than the da and db changes. I then read that the change has to be relative to the value or something like that so instead of using dc = 2 degrees, I did dc = sin(2) = 0.0349, which in fact does give me the correct answer.

My question: when determining that you can substitute the change in measured values for the dx variable, how do you know how to properly relate that change with the equation? Is it calculated by the expressed change in the function? For example, when you change the measured side by 1 ft, you are calculating 1 ft into the function, but when you change by 1 degree you are calculating sin(x+1) in the function. I'm trying to understand this better, thanks!

 

[Mark44]

Homework Statement

[/B]
The area of a triangle is (1/2)absin(c) where a and b are the lengths of the two sides of the triangle and c is the angle between. In surveying some land, a, b, and c are measured to be 150ft, 200ft, and 60 degrees. By how much could your area calculation be in error if your values for a and b are off by half a foot each and your measurement of c off by 2 degrees? See figure below

Homework Equations

[/B]
dA(x,y,z) = dA/dx dx + dA/dy dy + dA/dz dz
A = 1/2*absin(c)
a = 150+-0.5 ft, b = 200+-0.5 ft, c = 60+-2 degrees

The Attempt at a Solution

[/B]
I first used the chain rule on the Area to get dA

I don't see that you used the chain rule here. You calculated the differential of A, and used the product rule to get each of the three partial derivatives that make up the differential.

dA = 0.5bsin(c)*da + 0.5asin(c)*db + 0.5abcos(c)*dc

Setting da = 0.5, db = 0.5, and dc = 2

The reason your answer is off is that you are using degrees for c and dc. If you change the 2° to the equivalent radian measure, you should get the right answer.

and plugging in my values I get a completely wrong answer (the actual answer is 338 ft^2). Something felt off about the dc, because it varied so differently than the da and db changes. I then read that the change has to be relative to the value or something like that so instead of using dc = 2 degrees, I did dc = sin(2) = 0.0349, which in fact does give me the correct answer.

My question: when determining that you can substitute the change in measured values for the dx variable, how do you know how to properly relate that change with the equation? Is it calculated by the expressed change in the function? For example, when you change the measured side by 1 ft, you are calculating 1 ft into the function, but when you change by 1 degree you are calculating sin(x+1) in the function.

No, that's not what you do.
You use the given values of a, b, and c (namely, 150, 200, and $\pi/3$, respectively) and the given values of $\Delta a$, $\Delta b$, and $\Delta c$. The differential is a function of these six values.

I'm trying to understand this better, thanks!

 

[Nikstykal]

Why do you have to convert everything to radians to get the correct answer? Is that just the way trig functions work properly? Isn't using degrees and then using sin(2) as the change the equivalent?

 

[Mark44]

Why do you have to convert everything to radians to get the correct answer?

Because all of the numbers involved have to be real numbers, and radians are real numbers. Unlike degrees, radians are not really units.

Is that just the way trig functions work properly? Isn't using degrees and then using sin(2) as the change the equivalent?

You'll get the same results for $\sin(60°)$ and $\sin(\pi/3)$ it's true, but as you found, you get very different results if you use degree measure for $\Delta c$ instead of real numbers (i.e., radians).

 

[SteamKing]

Why do you have to convert everything to radians to get the correct answer? Is that just the way trig functions work properly? Isn't using degrees and then using sin(2) as the change the equivalent?

It's because for the derivatives (and integrals) of trig functions, the argument θ (the angle) is implicitly assumed to be measured in radians.
That's some of the fine print in the basic calculus of trig functions which you should have learned.

If θ is measured in degrees, for example, then d (sin (θ)) / dθ ≠ cos (θ), etc.

 

[Nikstykal]

Awesome, really appreciate the help. Thanks!