Derivative of L-sqrt(L^2-x^2)How do I take the derivative of L-sqrt(L^2-x^2)?

[amanda.ka]

Homework Statement

I did an experiment to test the conservation of mechanical energy in an oscillating pendulum. As part of the analysis I had to find the pendulum's vertical position with time using the formula: y = L-sqrt(L^2-x^2) where L was the pendulum's length (L=1 m). Then for the next step I had to find the vertical velocity vs. time. The instructions say that it can be found using derivative(y) so my question is how do I take the derivative of L-sqrt(L^2-x^2)?

Homework Equations

The Attempt at a Solution

When I tried to derive it I got:
f(x) = L - sqrt(L^2 - x^2) = L - (L^2 - x^2)^(1/2)
{d/dx}[f(x)] = f'(x) = - (1/2)[(L^2 - x^2)^(-1/2)][- 2x]
= x(L^2 - x^2)^(-1/2).

I think I did something wrong because when I put in the values of "x" I do not get the same answers as the computer program I am using to generate the graphs did. I have to show a sample calculation of how the data for the graph was calculated so any help would be appreciated, thanks!

 

[Svein]

Velocity = distance/time, so v_y = dy(t)/dt. You have y as a function of x, but you need y as a function of t...

 

[amanda.ka]

Velocity = distance/time, so v_y = dy(t)/dt. You have y as a function of x, but you need y as a function of t...

v = vo+at but I don't have acceleration, I am not sure how to make it just velocity as a function of time

 

[rude man]

Homework Statement

I did an experiment to test the conservation of mechanical energy in an oscillating pendulum. As part of the analysis I had to find the pendulum's vertical position with time using the formula: y = L-sqrt(L^2-x^2) where L was the pendulum's length (L=1 m). Then for the next step I had to find the vertical velocity vs. time. The instructions say that it can be found using derivative(y) so my question is how do I take the derivative of L-sqrt(L^2-x^2)?

Homework Equations

The Attempt at a Solution

When I tried to derive it I got:
f(x) = L - sqrt(L^2 - x^2) = L - (L^2 - x^2)^(1/2)
{d/dx}[f(x)] = f'(x) = - (1/2)[(L^2 - x^2)^(-1/2)][- 2x]
= x(L^2 - x^2)^(-1/2).

I think I did something wrong because when I put in the values of "x" I do not get the same answers as the computer program I am using to generate the graphs did. I have to show a sample calculation of how the data for the graph was calculated so any help would be appreciated, thanks!

You want dy/dt, not dy/dx. I hope you have dy/dx somewhere? x is of course related to L and θ, and you presumably have solved for θ(t)?

 

[amanda.ka]

You want dy/dt, not dy/dx. I hope you have dy/dx somewhere? x is of course related to L and θ, and you presumably have solved for θ(t)?

No i haven't found the angle θ, but I could find that with the pythagorean theorem from the diagram I'm assuming? I'm not sure how the angle plays into finding velocity as a function of time :/

 

[rude man]

No i haven't found the angle θ, but I could find that with the pythagorean theorem from the diagram I'm assuming? I'm not sure how the angle plays into finding velocity as a function of time :/

θ is the time-dependent variable, so it's really θ(t). To get θ(t) you either have to make lab measurements, which you seem to be doing, or solve the relevant differential torque equation relating rotational inertia I, angular acceleration θ'', and sum of torques τ :
I θ'' = Στ. To verify your lab measurements the latter is highly recommended in any case.
Vertical velocity d/dt (L-x) can then be related to dθ/dt since you can relate θ to x per Pythagoras.

I guess I'm not sure exactly what you're taking measurements of. Is it x? Or θ? Or ... ?

 

[amanda.ka]

θ is the time-dependent variable, so it's really θ(t). To get θ(t) you either have to make lab measurements, which you seem to be doing, or solve the relevant differential torque equation relating rotational inertia I, angular acceleration θ'', and sum of torques τ :
I θ'' = Στ. To verify your lab measurements the latter is highly recommended in any case.
Vertical velocity d/dt (L-x) can then be related to dθ/dt since you can relate θ to x per Pythagoras.

I guess I'm not sure exactly what you're taking measurements of. Is it x? Or θ? Or ... ?

We haven't covered torque yet but using a motion sensor I collected data for the horizontal displacement of the ball from the equilibrium position (x vs. time) then to find the vertical position vs. time I used the equation y= L-sqrt(L^2-x^2) where x referred to the values obtained from the x vs. time trials. Now I have to find the vertical velocity found from (dy/dt) the rate of change of the vertical position with time and I'm not sure how to go about doing that.

 

[Svein]

So you have made a pendulum (https://en.wikipedia.org/wiki/Pendulum_(mathematics)). For small displacements, the pendulum equation is: [itex]\theta (t)=\theta_{0}\cos(\sqrt{\frac{g}{L}t)} [/itex]. Now [itex]x=L\sin(\theta) [/itex], which means that [itex] \theta = \arcsin(\frac{x}{L})[/itex]. From that you can find [itex]\theta_{0} [/itex]. L and g you already know. This will give you ##x(t)## and through that ##y(t)##.

 

[rude man]

We haven't covered torque yet but using a motion sensor I collected data for the horizontal displacement of the ball from the equilibrium position (x vs. time) then to find the vertical position vs. time I used the equation y= L-sqrt(L^2-x^2) where x referred to the values obtained from the x vs. time trials. Now I have to find the vertical velocity found from (dy/dt) the rate of change of the vertical position with time and I'm not sure how to go about doing that.

OH, OK, I think I see . I thought x was = L - y. So x is the horizontal displacement and you have a set of data x(t), right?
Your equation y = L - sqrt(L^2-x^2) is of course right. This is just freshman calculus then: dy/dt = d/dt {L - sqrt(L^2-x^2)} and you have x(t) so you can get x'(t) from adjacent x data points, knowing the time between samples of x.

Forget about theta, it's not relevant to what you're trying to do as I see it.

 

[amanda.ka]

OH, OK, I think I see . I thought x was = L - y. So x is the horizontal displacement and you have a set of data x(t), right?
Your equation y = L - sqrt(L^2-x^2) is of course right. This is just freshman calculus then: dy/dt = d/dt {L - sqrt(L^2-x^2)} and you have x(t) so you can get x'(t) from adjacent x data points, knowing the time between samples of x.

Forget about theta, it's not relevant to what you're trying to do as I see it.

Could you please explain how I could get x'(t) from the adjacent x data points knowing the time between samples? Sorry, I haven't taken calculus yet so that's why I'm struggling with these basic concepts.

 

[rude man]

Could you please explain how I could get x'(t) from the adjacent x data points knowing the time between samples? Sorry, I haven't taken calculus yet so that's why I'm struggling with these basic concepts.

As a hint, an anlogy: suppose you weighed 150 lbs one day and, 24 hrs. later, you weighed 152 lbs. What would be your computed rate of change of weight gain?

 

[rude man]

As a hint, an anlogy: suppose you weighed 150 lbs one day and, 24 hrs. later, you weighed 152 lbs. What would be your computed rate of change of weight gain?

EDIT: OK, you don't need calculus. For every x data point there is a corresponding y data point. So to calculate rate of change of y (which is vertical velocity), just do the same with the y set of data points that you did with the x data points to compute horizontal velocity.

 

[amanda.ka]

EDIT: OK, you don't need calculus. For every x data point there is a corresponding y data point. So to calculate rate of change of y (which is vertical velocity), just do the same with the y set of data points that you did with the x data points to compute horizontal velocity.

I took y2-y1/t2-t1 and I got the right answer, thank you for your help!

 

[rude man]

I took y2-y1/t2-t1 and I got the right answer, thank you for your help!

OK! To improve accuracy you could compute (y3 - y1)/(t3 - t1) which would average the rate before and after t2.