# Derivative help

1. Dec 28, 2005

### tman1

i was given a problem and found that
x = (VcosA)/g*(VsinA + ((VsinA)^2 + 2gYo)^1/2)
where x is the distance
v is the speed
A is the angle
g is the force of gravity
and Yo is the height
but i am a bit lost in taking its derivative. once i do so, i need to maximize and minimize the angle. i was wondering if i could get some help

2. Dec 28, 2005

### saltydog

Hey Tman, first write it nicely. Yea, I know you don't know LaTex yet but the prettier it is the fewer the mistakes. That's just how it is in math:

Looks like:

$$\frac{vCos(a)}{g}\left[vSin(a)+\left(v^2 Sin^2(a)+2gy_0\right)^{1/2}\right]$$

Now, use the chain rule:

$$d(uv)=udv+vdu$$

right?

I'll start it for you:

$$\frac{vCos(a)}{g}\frac{d}{da}\left[vSin(a)+\left(v^2 Sin^2(a)+2gy_0\right)^{1/2}\right]$$

Last edited: Dec 28, 2005
3. Dec 28, 2005

### tman1

little more help

i understand that i must use the chain rule, but i dont know which terms are constants, for example vcosA and 2gYo. would i have to make them equal zero because of constant rule or no

4. Dec 28, 2005

### saltydog

Tell you what, how about this: I don't know the details of your problem but for now, just assume everything other than 'a' is constant. Now, take the derivative of the expression with respect to a and work it though and see what happens.

So continuing:

$$\frac{vCos(a)}{g}\left[vCos(a)+1/2\left(v^2Sin^2(a)+2gy_0)^{-1/2}(2v^2Sin(a)Cos(a))\right)\right]$$

5. Dec 28, 2005

### tman1

i think i need alittle more guidance

ok so from this point where i have the derivative of what u gave me, i can set it equal to zero and find the maximum for the angle, or is there more that i still need to do(ie more deriving). also it would help me alot if u cud tell me where u got ur trig identies from (ie there derivatives) so i can later retry this by myself.

6. Dec 28, 2005

### saltydog

That was just half of the derivative.

Tman . . . I don't wish to add complexity to this for you. So start with:

$$x=f(a)=\frac{vCos(a)}{g}\left[vSin(a)+\left(v^2 Sin^2(a)+2gy_0\right)^{1/2}\right]$$

and calculate the derivative (assuming all other parameters are constants):

\begin{align*} \frac{dx}{da}&=\frac{vCos(a)}{g}\left[vCos(a)+ 1/2\left(v^2Sin^2(a)+2gy_0)^{-1/2}(2v^2Sin(a)Cos(a))\right)\right] \\ &-\left[vSin(a)+\left(v^2Sin^2(a)+2gy_0\right)^{1/2}\right]\frac{vSin(a)}{g} \end{align}

That's just the chain rule and the basic rules for differentiation for sine and cosine and other functions.

Setting that to zero and solving for a, gives an extremum which could be a maximum or minimum. However that looks difficult to solve for.

However, this does not address the specific application you're working on (like shooting a projectile in the air with an initial velocity v, initial height y0, and angle a), and then determining the angle which gives the greatest distance. Not sure though this equation could be applied to that setup.

Last edited: Dec 28, 2005
7. Dec 28, 2005

### arildno

To OP:
Derivatives WITH RESPECT TO WHAT?? With respect to the equality sign, perhaps?

You have to learn to state your problems precisely, otherwise you'll end up not understanding them properly.