first derivative?....a little confused from a (sin),..is chain rule used??

Question

find the first derivative of R = g^-1 ((vi^2)sin(2(theta))) with respect to theta. Then if you can, show that by setting the first derivative to zero, 45 degrees = theta for maximum range.

im just really confused with the rules... plz help

Answer 1

The derivative of sin(2theta) is 2cos(2theta)

Yes, the chain rule apply because you want to differentiate sin(2theta) not sin(theta). Differentiate it by taking the 2theta as a whole first, which will give you cos(2theta), then differentiate what's inside (2theta)... which will yield 2. So the result is 2cos(2theta)

Then you have to apply the power rule twice to get the rest.

Answer 2

Are you sure you copied the problem correctly? Is vi^2 the same as (pi)^2 ? And you have only defined the argument for the function f^-1, not the function.

I think you meant to write the equation in polar form:

R = (pi)^2 sin(2T) for 0 <= T <= 2pi where T=theta

Then dR/dT is (pi)^2 times the derivative of sin(2T) times the derivative of 2T (by the chain rule, or

R' = (pi)^2 [cos(2T)] (2) = 4(pi)^2 cos(2T)

which equals zero when cos(2T) = 0, but cos(2T) is zero (x-coordinate on unit circle) at 2T = pi/2 and 3pi/2
T = pi/4, 3pi/4

The original function R= (pi)^2 sin(2T) .

Bu substitution and evaluation you will find it is a maximum at T=pi/4 = 45 degrees and a minimum at 3pi/4= 270 degrees

Answer 3

d(sin 2theta) = 2 cos (2 theta)

when 2cos(2 theta) = 0

cos( 2 theta) = 0

2 theta = pi/2

theta = pi/4 = 45 degrees

Answer 4

what's the vi???