## inkyvoyd 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

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

@hartnn , can you help me please?

3. inkyvoyd

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

4. hartnn

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

5. inkyvoyd

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

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

7. inkyvoyd

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

8. hartnn

oh, yes, sorry....

9. inkyvoyd

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

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

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

12. inkyvoyd

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

13. hartnn

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

14. inkyvoyd

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

15. hartnn

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

16. inkyvoyd

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

17. hartnn

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

18. inkyvoyd

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

that equality doesn't hold!

20. inkyvoyd

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

21. inkyvoyd

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

you know the right thing,thats enough

23. inkyvoyd

my grade is hurt though >.<

24. inkyvoyd

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

25. hartnn

hmm...welcome ^_^

26. amistre64

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

product rule for y'' ?