## inkyvoyd Group Title Definition of acceleration (calculus) "The position of an object is described by the parametric equations x=lnt and y=5t^2. What is the acceleration of the object in m/sec2 when t=2?" one year ago one year ago

1. inkyvoyd Group Title

I have gotten this problem wrong for calculating $$\sqrt{\frac{d^2y}{dt^2}+\frac{d^2x}{dt^2}}$$ instead of $$\frac{d^2y}{dx^2}$$ What am I doing wrong?

2. inkyvoyd Group Title

@hartnn , can you help me please?

3. inkyvoyd Group Title

I just need to know what I should be calculating for acceleration, and why. thanks!

4. hartnn Group Title

i would have calculated d^2y/dx^2 because i didn't know the sqrt formula :P

5. inkyvoyd Group Title

But, d^2y/dx^2 is the movement of y with respect to x - it doesn't say the particle is accelerating does it?

6. hartnn Group Title

if x is position, dy/dx gives velocity and d2y/dx2 gives the acceleration. so, it does.

7. inkyvoyd Group Title

no, dx/dt gives velocity - velocity is the derivative of distance with respect to time right?

8. hartnn Group Title

oh, yes, sorry....

9. inkyvoyd Group Title

I mean, the units don't add up when you do d^2y/dx^2 if you have y and x in parametric terms of t >.<

10. inkyvoyd Group Title

I've tried to email my teacher about it like 3 times but I don't think she's read what I've said - she just keeps repeating that the acceleration is d^2y/dx^2

11. hartnn Group Title

i am sure , just realised , that d2y/dx2 is NOT acceleration

12. inkyvoyd Group Title

Okay - how should I prove to my teacher that d^2y/dx^2 is NOT acceleration?

13. hartnn Group Title

units.....x and y are distances..... d2y/dx2 is dimensionless. it doesn't have the unit of acceleration.

14. inkyvoyd Group Title

Okay - any other ways? I tried emailing her saying units don't add up but I'm not sure if she read. >.<

15. hartnn Group Title

d2y/dx2 gives change in dy/dx w.r.t x which is in no way, acceleration

16. inkyvoyd Group Title

because the change in y with respect to x isn't velocity right?

17. hartnn Group Title

yeah, it isn't velocity , its just the slope

18. inkyvoyd Group Title

So then my teacher argues that d^2y/dx^2=$$\frac{\frac{d}{dt}(\frac{dx}{dt})}{\frac{dx}{dt}}$$ Which unfortunately has a d/t in it >.<

19. hartnn Group Title

that equality doesn't hold!

20. inkyvoyd Group Title

it does, but it has nothing to do with the definition of acceleration in two dimensions

21. inkyvoyd Group Title

yet I'm afraid my teacher will attempt to end the discussion and get back to doing other stuff (it's an online course) by simply telling me to ignore units.

22. hartnn Group Title

you know the right thing,thats enough

23. inkyvoyd Group Title

my grade is hurt though >.<

24. inkyvoyd Group Title

Ah well I'll think of something - thank you very much!

25. hartnn Group Title

hmm...welcome ^_^

26. amistre64 Group Title

spose we equate t in terms of x x=lnt ; e^x = t sub in e^x into the t for y y = 5e^(x^2) y' = 10x e^(x^2) y'' = 20x^2 e^(x^2) when t=2, x=ln2 y'' = 20(ln2)^2 e^(ln2^2) = 15.536 or so ....if my idea is correct

27. hartnn Group Title

product rule for y'' ?