# Maximum and Range of the Equation (Calc.)

1. Sep 15, 2011

### Calam1tous

1. The problem statement, all variables and given/known data

Hello,

I ran into this problem in the middle of my physics homework:

Using calculus, you can find a function’s maximum or minimum by differentiating and setting the result to zero. Do this for equation y = (u^2 / g)*sin(2x), differentiating with respect to x, and thus find the maximum range for x.

u = initial velocity
g= acceleration of gravity
x = theta

2. Relevant equations

Possibly x = ut for range?

3. The attempt at a solution

It's been a several months since I've done this type of problem, but I tried to differentiate it through the quotient rule (and product rule) and got:

(2*g*u^2*cos(2x) + 2*g*u*sin(2x) - u^2*sin(2x)) / g^2

I tried to set this to zero in order to find the maximum:

0 = (2*g*u^2*cos(2x) + 2*g*u*sin(2x) - u^2*sin(2x)) / g^2

but I couldn't figure out what to do and I had a suspicion I was doing everything wrong.

Can anyone point out my mistakes / what to do next?
Thanks!

Last edited: Sep 15, 2011
2. Sep 15, 2011

### Staff: Mentor

It would appear that u and g are constants, so that (u2/g) is also a constant. So you only have to worry about differentiating the sin(2x).

What's the derivative of sin(2x) with respect to x?

3. Sep 16, 2011

### Calam1tous

Thanks for the input, treating the u and g variables as constants made the problem make more sense and I was able to solve it.

y = range
x = theta

Solution (for reference):

If I ignore the other variables and differentiate the equation, y = sin(2x), I get:

y' = 2cos(2x)

Then set it to 0 to find the critical point:

0 = 2cos(2x)

Divide by 2:

0 = cos(2x)

Take the inverse cosine of both sides:

cos^-1(0) = 2x

Divide both sides by 2:

90 / 2 = x

45 degrees = x

Therefore the maximum range given by x (theta) is 45.

4. Sep 16, 2011

### SammyS

Staff Emeritus
That gives the value of x that makes y a maximum.

The question asks for the maximum range, which is the y value when x = 45 degrees.