# Derivative of projectile parametric y-component?

• Matt Jacques
In summary, the conversation is about finding the derivative of the projectile parametric y-component. The y-component is given by the formula y = (vi/k)(1-e^-kt)(sin a) + (g/k^2)(1 - kt - e^-kt). The person is having trouble with their derivative and wants to check it against the final answer. The derivative is found to be dy/dt = Vie-ktsin(a)+g(e-kt-1)/k. Next, they discuss how to solve the derivative by setting it to zero and solving for t. They suggest expanding the second term and combining terms to isolate the exponential, then taking the natural log of both sides. Finally, the person thanks the other for
Matt Jacques
Hi,

I'm looking for the derivative of the projectile parametric y-component?

The y component is:

y = (vi/k)(1-e^-kt)(sin a) + (g/k^2)(1 - kt - e^-kt)

I seem to be doing something wrong and my derivative isn't working out, I just want to check it against the final answer to see where I am going wrong.

Thanks

Matt

I get
dy/dt = Vie-ktsin(a)+g(e-kt-1)/k

Yep, that's it. Thanks, Integral.

I forgot d/dx [ e^u ] is e^u * du/dx

Ah, the devil is in the details! Now to step two, how to solve it? I logged both sides, used the power rule and everything else I could think of, but it is not working out. Any other suggestions beside logging?

[Edited for spelling mistake]

No ideas? :(

What do you mean by "solve it"?? Presumably you don't mean "How did you get that answer" because you said "Yep, that's it". What exactly do you want to do with the derivative?

Ooops, sorry.

Im settng the derivative to zero and solving for t.

The first term contains an exp(-kt). Expand out the second term into two separate terms, one of which will contain the same exp(-kt) factor. Combine terms containing the exponential, move everything else over to the other side and once the exponential is isolated, then take the natural log of both sides.

I got something like

t = (-1/k)*ln[(g/k)*(Vi*sin(a)+g/k)^(-1)]

Last edited:
Thanks, Futz. I was combining the exponentials wrong.

Here it is solved for t:

http://homepage.mac.com/jjacques2/maxheight.jpg

Last edited by a moderator:

## 1. What is the formula for the derivative of the y-component of a projectile's parametric equation?

The formula for the derivative of the y-component (vertical motion) of a projectile's parametric equation is dy/dt = v*sin(theta) - g*t, where v is the initial velocity, theta is the launch angle, g is the acceleration due to gravity, and t is the time.

## 2. How is the derivative of the y-component of a projectile's parametric equation used in physics?

The derivative of the y-component is used to calculate the vertical velocity and acceleration of a projectile at any given time, which is important in understanding the motion of the projectile in physics.

## 3. Can you explain the significance of the launch angle in the derivative of a projectile's parametric equation?

The launch angle, theta, affects the initial vertical velocity of a projectile and therefore impacts the shape and trajectory of its path. It is a key component in calculating the derivative of the y-component of a projectile's parametric equation.

## 4. Is the derivative of the y-component of a projectile's parametric equation affected by air resistance?

Yes, the derivative is affected by air resistance because it is a force that acts in the opposite direction of the projectile's motion and can decrease its vertical velocity over time.

## 5. How does the derivative of the y-component of a projectile's parametric equation change at the highest point of its trajectory?

At the highest point of its trajectory, the derivative of the y-component is equal to zero, as the vertical velocity is momentarily zero before the projectile begins to descend due to gravity.

Replies
4
Views
1K
Replies
4
Views
1K
Replies
4
Views
2K
Replies
6
Views
1K
Replies
6
Views
689
Replies
14
Views
2K
Replies
1
Views
3K
Replies
2
Views
780
Replies
25
Views
2K
Replies
7
Views
1K