Hi, I have a problem and I am stuck to one question: The area of a triangle is given by the formula: A=1/2*b*c*sin(θ ) At time to, we have bo=10 inches, co=20inches, θo=pi/3 a) Find the area of the triangle at to I found 50*sqrt(3) b- find the rate of change of the area with respect to b at to. I found 5*sqrt(3). c) Find the rate of change of the area with respect with θ at to. I found 50. d) Using the rate found in part c) , calculate (by differentials) the approximate change in area if angle theta is increased by one degree. I need help for this question. I do not know what to do here. Thank you for your help
You found the rate of change of area with respect to θ to be 50 sq. inches per radian (angles here measured in radians). This means that close to the given conditions (bo=10 in., co=20in., θo=pi/3 radians) a "small" change of 1 radian in the triangle's included angle θo changes its area by 50 square inches. (Note: "Differentials" are very small changes - the smaller the change, the more accurate the approximation will be. A change of 1 radian is very big, almost 60 degrees - so that wouldn't give a reasonable answer. But a differential of 1 degree is fine, depending on how accurate you want your answer to be.) So, close to the initial dimensions, the change in area (measured in sq. inches) is 50 x the change in angle θo (measured in radians). The hardest part here is converting the angle to radians (multiply by 2pi and divide by 360).