# Partial derivations

Joined Dec 29, 2004
83
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.

#### barrygale2003

Joined Oct 11, 2005
2
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).

Originally posted by braddy@Oct 9 2005, 02:47 PM
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

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.