# Derivatives in relation to physics

1. Oct 27, 2015

OK, I have never had physics till this semester and I am in calculus based physics and it is kicking my butt.

I don't understand how derivatives are properly used in the formulas, and I have an example, my question is the image attached.

@=theta
A=alpha

SO,
A) I need omega (angular velocity). w= d@/dt right? so if @=0.11(4.9)^2, plugged into that equation...
we end up with:
w=d(0.11*4.9^2)/d(4.9)
w=0.11(2)(4.9)/1=1.08 rad/s, <------this is right I think.

B) got it: v=rw, no sweat, v=10.8m/s

C) Here is where I trip up. According to my understanding, A=dw/dt, which means d(1.08)/d(4.9)....

but thats not right, derivatives of constants=0....I don't understand. It's been a while since I took calculus, so am I just messing something dumb up???

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2. Oct 27, 2015

and to boot, the formula for omega(w) is just listed as w=@/t, which gives me 0.539 rad/s.

3. Oct 27, 2015

is it because they are differential change of @ and t? sooooo, its actually d((0.11*t^2)-0)/d(4.9-0) so the differential just takes the 0 away?

...no that isn't right.

4. Oct 27, 2015

### SteamKing

Staff Emeritus
Maybe you should refresh your knowledge of calculus before trying to apply it to physics.

Back to the problem at hand:

θ = 0.11 t2

If you want to find the angular velocity ω = dθ / dt, then differentiate the expression for θ with respect to t first, before plugging in any values for t.

https://www.mathsisfun.com/calculus/derivatives-rules.html

What you wrote, w=d(0.11*4.9^2)/d(4.9), made no sense whatsoever. The derivative, dθ / dt, does not mean substitute the expression for θ, the value for t, and cancel out the d's. It means something altogether different.

5. Oct 27, 2015

So it is in leibniz notation, that explains a lot of the issues I am having.

So Id I write it this way, I end up with d/dt (0.112t) = 1.08 @ t=4.9s

but regardless, I still don't understand when I find angular acceleration why it is the dw/dt when to get the answer I don't take a derivative anywhere. I just divide w by t. Is it because it is actually the second derivative of d@/dt so I end up with 0.11(2) =0.22 which is the answer....yuuup

Please excuse my inabilities with leibniz, My professor stuck to newton notation, with y dots or f'(x). He touched on leibniz notation, but I never quite understood it. Since I hadn't planned on going further I didn't bother to pick it up. I will have to remedy that as it evidently starting to cause me problems.

6. Oct 27, 2015

I guess my issue here is, how do I know when I am using an equation what level of derivative I am actually at? Is it just something I have to know?

7. Oct 27, 2015

### Staff: Mentor

You need to go back and review calculus. These are all math issues that you're asking about, not Physics. These are typical homework problems covered in a calculus course.

Chet

8. Oct 27, 2015

### SteamKing

Staff Emeritus
No, that's still not correct.

If θ =0.11 t2, then dθ / dt = 2 * 0.11 * t = 0.22 t, for all values of t.

If you want to find the value of the angular velocity at a particular time t0, then dθ / dt = 0.22 t is evaluated by setting t = t0, thus

At t0 =4.9 s, dθ / dt = 0.22 * 4.9 = 1.08 rad/sec = angular velocity @ t = 4.9 sec

The angular acceleration is defined as the change of angular velocity with respect to time. Since angular velocity = dθ / dt = ω, then angular acceleration α = dω / dt.
For this particular example, ω = 0.22 t, so α = 0.22 rad / sec2, which is a constant value for any time t = t0.
The two forms of derivative notation are slightly different, but the concepts are the same:

$dθ / dt = \dot{θ}$
$dω / dt = \dot{ω} = \ddot{θ}$

Each dot above a variable indicates differentiation with respect to time. More than one dot indicates a higher-order time derivative.

As Chet indicated above, your problem is with calculus, rather than physics. Spend some time learning (or re-learning) the calculus, and then come back to the physics.

9. Oct 27, 2015