Maximum and Range of the Equation (Calc.)

[Calam1tous]

Homework Statement

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

Homework Equations

Possibly x = ut for range?

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!

 

[gneill]

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?

 

[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.

 

[SammyS]

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.

 

[rwisz]

Hello,

You are on the right track with using the derivative to find the maximum value of a function. However, in this case, we are looking for the maximum range for x, not the maximum value of the function itself. To find the maximum range, we need to find the value of x that gives the maximum value for y.

To do this, we can set the derivative equal to zero and solve for x:

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

We can then simplify this equation to:

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

Next, we can factor out a 2u from the first two terms and a u from the last term:

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

Now, we can set each factor equal to zero and solve for x. This will give us two values for x, one for the maximum range and one for the minimum range.

First factor:

2u = 0, so u = 0

This means that when the initial velocity is 0, the range is also 0. This makes sense because if there is no initial velocity, the object will not travel any distance.

Second factor:

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

We can rearrange this to:

u(cos(2x) - sin(2x)) = -sin(2x)

Dividing both sides by u gives us:

cos(2x) - sin(2x) = -1/u

Using trigonometric identities, we can rewrite this as:

√2*cos(2x - π/4) = -1/u

Solving for x gives us:

x = (π/8 + nπ/2) / 2, where n is any integer

We can plug these values of x back into the original equation to find the maximum range for each value of n. The maximum range will depend on the initial velocity, u, and the acceleration of gravity, g. So, the maximum range for x will be:

x_max = (π/8 + nπ/2) / 2