Partial Drivative- Inclined Plane Problem

[ritwik06]

Homework Statement

A ping pong ball is projected (with velocity v)from a foot of an inclined plane with an inclination of \beta . The initial velocity of projection makes an angle \alpha with the surface of th incline. Find the relation beteen \alpha and \beta such that the range is maximum.

The Attempt at a Solution

Time of flight=\frac{2v sin \alpha}{g cos \beta}

Range(R)=\frac{2 v^{2} sin \alpha cos \alpha}{g cos \beta}-\frac{4 tan \beta v^{2} sin ^{2} \alpha}{g cos \beta}

Now the range is a function of both \alpha and \beta. I have been searching how to do the derivative. I learned from mathworld.wolfram that I need to consider one quantity constant and differentiate with respect to the other. and then do the same by keeping the other constant. Then by adding these partial derivatives I can get the complete derivative of the function concerned (which in my case is the range of the projectile).

\frac{\delta R}{d\alpha}=\frac{2 v^{2} cos 2\alpha}{g cos \beta}-\frac{4 v^{2} tan \beta sin 2\alpha}{g cos \beta}


\frac{\delta R}{d\beta}=\frac{v^{2} sin 2\alpha tan \beta}{g cos \beta}-\frac{4 v^{2} sin^{2} \alpha(2tan^{2}\beta+1)}{g cos \beta}

Adding these two and putting them equal to zero does not yield the desired result. I am stuck. Please help me with this.

 

[ShellSnail]

One method is to resolve the velocity in the plane parallel to the incline slope and perpendicular to the inclined slope.

wait while i try it out

 

[ritwik06]

One method is to resolve the velocity in the plane parallel to the incline slope and perpendicular to the inclined slope.

wait while i try it out

I done just that. I have resolved it along the surface of incline and perpendicular to it.

Yup! take your time!

 

[ShellSnail]

x = ucos\betat - 0.5gsin\alphat^{2} ----1
y = usin\betat - 0.5gcos\alphat^{2} ---- 2
At the point of impact, y = 0
Hence, t = (2usin\beta) / (gcos\alpha) --- 3

put 3 into 1
and differentiate with respect to \beta

you shld end up with

tan 2\beta = cot\alpha

thus, \beta = pi/4 - \alpha/2

EDIT: switch all the alpha with beta and beta with alpha

 

[gabbagabbahey]

You have an error in your Range; you should have

R=\frac{2 v^{2} sin \alpha cos \alpha}{g cos \beta}+\frac{2 tan \beta v^{2} sin ^{2} \alpha}{g cos \beta}

It looks like you forgot to multiply your at^2 term by (1/2), and it should be +(1/2)at^2, not minus. (Assuming your g is -9.8m/s^2)

The total derivative of R is given by:

dR=\frac{\partial R}{\partial \alpha}d \alpha +\frac{\partial R}{\partial \beta}d \beta

At a local max/min the total derivative will be zero, so each partial derivative must also vanish. Compute the partial derivatives and set them equal to zero.

 

[ritwik06]

x = ucos\betat - 0.5gsin\alphat^{2} ----1
y = usin\betat - 0.5gcos\alphat^{2} ---- 2
At the point of impact, y = 0
Hence, t = (2usin\beta) / (gcos\alpha) --- 3

put 3 into 1
and differentiate with respect to \beta

you shld end up with

tan 2\beta = cot\alpha

thus, \beta = pi/4 - \alpha/2

EDIT: switch all the alpha with beta and beta with alpha

Can you please explain how you got that relation?

 

[gabbagabbahey]

Hmmm...did you understand my last post? Is there anything in there that you don't understand?

 

[ritwik06]

Hmmm...did you understand my last post? Is there anything in there that you don't understand?

To an extent...
You were right that i forgot to multiply that by 1/2 but there would be a minus sign there.

I didn't understand how you got the total derivative?
And then is it necessary that if z=x+y and z=0, then how come can we say that x=0, and y=0 are the only solutions?

And I am facing problems in deriving the required relation between alpha and beta!

 

[gabbagabbahey]

The total derivative of a function of two variables, say f(x,y) is by definition:

df=\left| \frac{\partial f}{\partial x} \right| d x +\left| \frac{\partial f}{\partial y} \right| d y

Since x and y are independent variables, so are there total derivatives dx and dy; and so the only way that df can be zero is if each of the partial derivatives \frac{ \partial f}{\partial x} and \frac{ \partial f}{\partial y} are zero.

Do you follow this?

 

[ritwik06]

The total derivative of a function of two variables, say f(x,y) is by definition:

df=\left| \frac{\partial f}{\partial x} \right| d x +\left| \frac{\partial f}{\partial y} \right| d y

Since x and y are independent variables, so are there total derivatives dx and dy; and so the only way that df can be zero is if each of the partial derivatives \frac{ \partial f}{\partial x} and \frac{ \partial f}{\partial y} are zero.

Do you follow this?

Yup! I do it now. As x is independent of y at all times, therefore we cannot force it to become a function of y to get the minimum value. right?
just like saying if
x i+ y j=0 i+ 0 j
in vector notation, then x=0 and y=0. right >

 

[gabbagabbahey]

More or less, yes.

So in this case your function is R(\alpha,\beta) and so \frac{\partial R}{\partial \alpha} and \frac{\partial R}{\partial \beta} must both be zero at the local max/min...what do you get when you compute these partial derivatives...where are they zero?

 

[ritwik06]

SOLVED

More or less, yes.

So in this case your function is R(\alpha,\beta) and so \frac{\partial R}{\partial \alpha} and \frac{\partial R}{\partial \beta} must both be zero at the local max/min...what do you get when you compute these partial derivatives...where are they zero?

Thanks a lot! I am very grateful to you. I have got my result.

 

[gabbagabbahey]

Congrats!