Differential of triangles and anlges

[muso07]

Given cosine rule: L=$$\sqrt{(r_{1})^{2}+(r_{2})^{2}-2r_{1}r_{2}cosx}$$

Consider a triangle with side lengths measured at $$r_{1}=3, r_{2}=4$$, and included angle x=$$\pi/2$$, each measured accurate to within 1%. Write down the differential dL in terms of $$dr_{1}, dr_{2}$$ and $$dx$$, and use this to estimate the maximum possible percentage error in L.

Any help? :S

 

[HallsofIvy]

What kind of help do you want? You haven't done anything at all. You are asked to write the differential dL. What is the derivative of L with respect to each of the variables? You might find it easier to use $L^2= r_1^2+ r_2^2- 2r_1r_2 cos(x)$ and find the differential from that.

 

[muso07]

I got dL= 1/L (0.03(r1-r2cosx) + 0.04(r2-r1cosx) + 0.005pi(r1r2sinx)).. wasn't sure if that was right.

 

[muso07]

My bad..
I got $$|dL|\leq\ 1/L[0.03(r_{1}+r_{2}cosx)+0.04(r_{2}+r_{1}cosx)-0.005\pi(r_{1}r_{2}sinx)]$$

If that's right, any hints on what I do next?