Solving for u & A: Help Appreciated!

[DivGradCurl]

Guys, I'm a little confused. Please, take a look at this:

$$L=\frac{u\cos A}{r} \left( 1 - e^{-rt} \right)$$

Solving for $$t$$ gives

$$t=\frac{1}{r}\ln \left( \frac{u\cos A}{u\cos A - rL} \right)$$

Then, substitute $$t$$ in the equation that follows

$$H=-\frac{gt}{r}+ \frac{1}{r}\left( u\sin A + \frac{g}{r} \right) \left( 1 - e^{-rt} \right) + h$$

which gives

$$H = \frac{g}{r}\ln \left( 1 - \frac{rL}{u\cos A} \right) + \frac{L}{u\cos A} \left( u\sin A + \frac{g}{r} \right) + h$$

Let's say we're given the values:

$$g=32$$

$$h=3$$

$$r=\frac{1}{5}$$

$$L=350$$

$$H=10$$

That implies we haven't yet obtained $$u$$ and $$A$$. There is ONE equation and TWO variables. However, we're looking for the $$u_{min}$$ AND $$A_{max}$$. I think we first need to solve the equation above for $$u$$, take the first derivative of the expression with the variable $$A$$, set it equal to zero, and then solve it for $$A_{max}$$. Consequently, we're are able to get $$u_{min}$$.

I've had difficulty solving the equation for $$u$$. I also tried to solving it with aid of the computer, but it won't give me the answer!

Any help is highly appreciated.

 

[OlderDan]

First, check your substitution. It does not look right to me. Either that or your H equation is wrong to begin with.

 

[thepatientmental]

Hi there,

Thank you for sharing your question and confusion with us. It looks like you are trying to solve for u and A in the equation provided. This can be a bit tricky since there are two variables and only one equation. However, you are on the right track with using the given values to solve for these variables.

To solve for u, you can rearrange the equation for t and substitute the given values to get:

t = 1/5 * ln(u*cosA/(u*cosA - 1750))

Next, substitute this value for t into the equation for H and set it equal to the given value of 10. This will give you an equation with only u and A as variables. You can then use a computer or a graphing calculator to find the values of u and A that satisfy this equation.

As for finding u_min and A_max, you are correct in thinking that you will need to take the first derivative of the equation for H with respect to A and set it equal to 0 to find the maximum value of A. Once you have this value, you can use it to solve for u_min.

I hope this helps. If you are still having difficulty, it may be helpful to discuss the problem with a tutor or teacher who can provide more guidance. Good luck!