Boy scout problem (Serway and Jewett)

[Kavya Chopra]

Homework Statement

To protect his food from hungry bears, a boy scout raises his food pack with a rope that
is thrown over a tree limb at height $$h$$ above his hands. He walks away from the vertical rope
with constant velocity $$v_b$$, holding the free end of the rope in his hands
(a) Show that the speed $$v$$ of the food pack is given by $$x(x^2 + h^2)^\frac{–1}{2} v_b$$ where $x$ is the distance he has walked away from the vertical rope.
(b) Show that the acceleration a of the food pack is $$h^2(x^2 + h^2)^\frac{-3}{2} v_b^2$$

I've done a), and am having a problem with b)

Homework Equations

$$v\frac{dv}{dx}=a$$

The Attempt at a Solution

I used the above equation to get
$$x(x^2 + h^2)^\frac{–1}{2} v_b\frac{d(x(x^2 + h^2)^\frac{–1}{2} v_b)}{dh}$$

Solving the partial derivative, I got
$$x(x^2 + h^2)^\frac{–1}{2} v_b hx(x^2 + h^2)^\frac{–3}{2}v_b$$
Which doesn't match the answer.
So, where am I going wrong?

Also, any alternative non-calculus solutions would be appreciated.

 

[haruspex]

$$v\frac{dv}{dx}=a$$.

Where x is what, in the context of the problem?

 

[Kavya Chopra]

Where x is what, in the context of the problem?

h, since the velocity and acceleration are both in the direction of the vertical distance h

 

[haruspex]

h, since the velocity and acceleration are both in the direction of the vertical distance h

But h is a constant. You cannot differentiate wrt a constant.
You need the distance of which v is the derivative wrt time.

 

[Kavya Chopra]

It isn't x, is it?
That's the only other variable distance on which velocity depends.
On the other hand, do we have to integrate velocity w.r.t. Time to get that distance?
To me the second one seems more appropriate, but then how to we find time?
Which one to do, and why?

 

[ehild]

He walks away from the vertical rope with constant velocity $$v_b$$, holding the free end of the rope in his hands
(a) Show that the speed $$v$$ of the food pack is given by $$x(x^2 + h^2)^\frac{–1}{2} v_b$$ where $x$ is the distance he has walked away from the vertical rope.
(b) Show that the acceleration a of the food pack is $$h^2(x^2 + h^2)^\frac{-3}{2} v_b^2$$

Homework Equations

$$v\frac{dv}{dx}=a$$

.

The last equation is not correct.It should be $$\frac{dx}{dt}\frac{dv}{dx}=a$$
What is $\frac{dx}{dt}$?
Why do you want to integrate velocity with respect time? How is acceleration defined?

 

[haruspex]

It isn't x, is it?

No, it isn't.

That's the only other variable distance on which velocity depends.

How about a function of h and x?
What is this velocity v? The rate of change of what distance?

The last equation is not correct.

Looks ok to me. The two v's must represent dx/dt, though.

 

[ehild]

Looks ok to me. The two v's must represent dx/dt, though.

v is the velocity of the food; vertical. It depends on the horizontal distance traveled by the boy.
v is function of x. The acceleration of the food is dv/dt=(dv/dx) (dx/dt). And dx/dt=vb, where vb is the velocity of the boy.

 

[haruspex]

v is the velocity of the food; vertical. It depends on the horizontal distance traveled by the boy.
v is function of x. The acceleration of the food is dv/dt=(dv/dx) (dx/dt). And dx/dt=vb, where vb is the velocity of the boy.

We are at cross purposes, but effectively saying the same thing.
I wrote that the equation a = v dv/dx is fine so long as meanings are such that v=dx/dt. Since the v in that equation is not the velocity of the boy, the x in that equation is not the displacement of the boy. See post #2.

But if we allow x to be an arbitrary parameter we can still write a = dv/dt = (dv/dx)(dx/dt).

You are taking x to be the displacement of the boy, as in the question statement, in which case, as you posted, dx/dt=v_b. I agree that looks like the simpler approach.

 

[ehild]

But if we allow x to be an arbitrary parameter we can still write a = dv/dt = (dv/dx)(dx/dt).

You are taking x to be the displacement of the boy, as in the question statement, in which case, as you posted, dx/dt=v_b. I agree that looks like the simpler approach.

Well, the problem definitely said that x is the displacement of the boy. So the OP-s equation a=v dv/dx was not correct.

 

[haruspex]

Well, the problem definitely said that x is the displacement of the boy. So the OP-s equation a=v dv/dx was not correct.

It was listed as a "relevant equation". That is, it is quoted from some notes or text. As such, the equation was fine, but, as ever, it ought to be accompanied with "... where a, v and x are ...".
A significant proportion of the errors on the forum are down to equations being used with the wrong association between formal and actual parameters.

 

[ehild]

It was listed as a "relevant equation". That is, it is quoted from some notes or text. As such, the equation was fine, but, as ever, it ought to be accompanied with "... where a, v and x are ...".
A significant proportion of the errors on the forum are down to equations being used with the wrong association between formal and actual parameters.

I agree.

 

[Kavya Chopra]

Yeah @above is right. I didn't explicitly mention what the variables were referring to. Here by x, I didn't mean the horizontal displacement in the equation, but the general form by which such an equation is represented.
What I meant was

$$\frac{dh}{dt}\frac{dv}{dh}=a$$

But as the above pointed out, you can't differentiate w.r.t. a constant.
So @above^2's approach of
$$\frac{dx}{dt}\frac{dv}{dx}=a$$
is the correct one since x is variable. That gives the correct answer.

What I was wondering was that why was my equation not correct. I thought that we can't use $$\frac{dv}{dx}$$ as they are both perpendicular to each other and hence have no components in each other's direction (forgetting that $$\frac{dv}{da}=t$$ is true for circular motion)

One last question: as long as A is a function of B, no matter what the angle between them is, can we differentiate or integrate A w.r.t. B?( A and B are physical quantities)
(Sorry, I'm just a beginner at calculus and calculus based physics, so I'm not very good at it)

 

[haruspex]

as long as A is a function of B, no matter what the angle between them is, can we differentiate or integrate A w.r.t. B?

Yes, and also if B is a function of A. In graphical terms, if you have a graph of y against x, ∫y.dx gives the area between the curve and the x axis, while ∫x.dy gives the area between the curve and the y axis.

 

[Kavya Chopra]

OK. Thanks a lot.