Thread: optimized problem

the two eqal sides of an isosceles triangle were each measured to have lengths 4 inches. the angle between these sides was measured to be pi/3 radians. The maximum error in the linear measurements is 0.02 inches, while the maximum error in the angle measure is 0.017 radians. Use differentials to approximates the maximum possible error in computing the area of the triangle with approriate units.

Originally Posted by ggw

the two eqal sides of an isosceles triangle were each measured to have lengths 4 inches. the angle between these sides was measured to be pi/3 radians. The maximum error in the linear measurements is 0.02 inches, while the maximum error in the angle measure is 0.017 radians. Use differentials to approximates the maximum possible error in computing the area of the triangle with approriate units.

Write the area of the isosceles triangle in terms of the length of the equal
sides and the angle:

A = x^2 cos(theta/2) sin(theta/2) = x^2 sin(theta)/2

Then:

delta(A) ~= [D_x A] delta(x) + [D_{theta} A] delta(theta)

where D_x is the partial differential operator with respect to x and
D_{theta} is the partial differential operator with respect to theta.

D_x A = x sin(theta)

D_{theta} A = x^2 cos(theta)/2

so:

delta(A) = x sin(theta) delta(x) + x^2 cos(theta)/2 delta(theta)

Now x=4 inches, delta(x)=0.02 inches, theta=pi/3 rad, delta(theta)=0.017 rad

so:

delta(A) = 4 sin(pi/3) 0.02 + 16 cos(pi/3)/2 0.017 ~= 0.137 sq inches

RonL