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

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.

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 $\displaystyle \frac{dF}{F}= \frac{udv+ vdu}{uv}= \frac{du}{u}+ \frac{dv}{v}$. 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 $\displaystyle \dfrac{dS}{S}= \dfrac{(1/2)(db c sin(A)+ b dc sin(A) + bc cos(A) dA}{bc sin(A)}= \dfrac{db}{b}+ \dfrac{dc}{c}+ \dfrac{cos(A)}{sin(A)}dA$

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".