# Math Help - Total Differential, percentage error

1. ## Total Differential, percentage error

I am having problems with the following question:

"The area of a triangle ABC is calculated using the formula:

S=1/2bcsinA

and it is known that b, c and A are measured correctly to within 1%. If the angle A is measured as 45 degrees, prove that the percentage error in the calculated value of S is not more than about 2.8%."

When I use the equation:

(delta)S= (dS/db)(delta b) + (dS/dc)(delta c) + (dS/dA)(delta A)

and subsequently divide by S, I get:

(delta)S/S= (delta)b/b + (delta)c/c + (cosA/sinA)(delta A)

and at this point I am totally stuck. What happens to the delta A here? And for the cosA/sinA do I use angle A as 45 degrees?

Any help would be fantastic.

Thanks

2. ## Re: Total Differential, percentage error

Hey lauramorrison93.

Since you know the angle, you can calculate the delta by using information about the relative error.

If something is off by 1% at the most then the delta will be 0.01*45*pi/180 (convert to radians) which is 0.007853 which is the value for delta(A) (in radians)

This means we dS/S = 0.01 + 0.01 + 0.007853 = 0.027853 < 0.028 which proves the claim made above.

3. ## Re: Total Differential, percentage error

There is, by the way, an engineer's "rule of thumb" that "when measurements are added, errors add. When measurements are multiplied, relative errors add. That is because if F= u+ v, dF= du+ dv while if F= uv, dF= udv+ vdu so that $\frac{dF}{F}= \frac{udv+ vdu}{uv}= \frac{du}{u}+ \frac{dv}{v}$. Here. we have S= (1/2)bc sin(A) so that dS= (1/2)(db c sin(A)+ b dc sin(A)+ bc d(sin(A)).
Of course, d(sin(A))= cos(A)dA so dS= (1/2)(db c sin(A)+ b dc sin(A)+ bc cos(A) dA).

Then the "relative error" is $\dfrac{dS}{S}= \dfrac{(1/2)(db c sin(A)+ b dc sin(A) + bc cos(A) dA}{bc sin(A)}= \dfrac{db}{b}+ \dfrac{dc}{c}+ \dfrac{cos(A)}{sin(A)}dA$
You do not need to know the measured values of b and c because they only occur linearly so you only need "db/b= dc/c= .01".