finding the minimum value of an equation with derivative?

Question

The range of a projectile fired with an initial velocity (v0) at an angle x (or theta) with the horizontal is
R=(v0sin2(x))/g, where g is acceleration due to gravity. find the angle x such that the range is a minimum.
Ok, well i took the derivative of it, (v0 and g are constants, so it would just be like taking the derivative of sin2x.
but then, im stuck.
i set it to zero, right, and then find those values.
but what i got for a derivative doesnt match that in the back of the book.
what is the correct derivative of this function?

Answer 1

well, here goes nothing. i have an answer, but i'm not 100% sure if it's right. as you said, (V0/g) is a constant. the derivitive of sin(2x) is 2cos(2x). set equal to 0 and i got:
(V0/g)2cos(2x)=0
divide both sides of equation by g/2V0:
cos(2x)=0
then i graphed the function of cos(x) and saw that it equalled 0 at (pi/2) and (3pi/2). we're looking for minimum so we keep (pi/2) and discard (3pi/2). since we're dealing with 2x and not x we have:
2x=(pi/2)
divide both sides by 2
x=(pi/4) or 45 degrees. hope this matches up to the back of the book

Answer 2

The derivative won't give you a minimum. Just look at the function sin(2x) for 0<=x<=pi/2 (or 90°)... It starts at 0, then is positive, maxes out at 1 for x=pi/4(45°) and goes down again to 0 for x=pi/2. If you graph it, it's clear that the zero derivative will show you the max but not the min.

It's also intuitively clear: Throw an object at a 90° angle (straight up) or at a 0° angle (horizontally from the ground) and it's going to travel 0 distance in the air.

Answer 3

R = (v0^2sin2θ)/g
dR/dθ = 2(v0^2/g)cos2θ = 0 for POSSIBLE max or min
2θ = 90°
θ = 45°,
but this yields a max, not a min, so take a look at the endpoints of the domain, 0° and 90°.
sin(2*0) = 0, and sin(2*90) = 0, which are your minimums.

(This was a particularly nasty trick question. You're expecting to have to differentiate, but that's not even necessary when you have a function with a 0 value that is constrained not to go negative.)