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$$\beta$$t - 0.5gsin$$\alpha$$t$$^{2}$$ ----1
y = usin$$\beta$$t - 0.5gcos$$\alpha$$t$$^{2}$$ ---- 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$$\beta$$t - 0.5gsin$$\alpha$$t$$^{2}$$ ----1
y = usin$$\beta$$t - 0.5gcos$$\alpha$$t$$^{2}$$ ---- 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!