inkyvoyd
  • 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?"
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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inkyvoyd
  • 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?
inkyvoyd
  • inkyvoyd
@hartnn , can you help me please?
inkyvoyd
  • inkyvoyd
I just need to know what I should be calculating for acceleration, and why. thanks!

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hartnn
  • hartnn
i would have calculated d^2y/dx^2 because i didn't know the sqrt formula :P
inkyvoyd
  • 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?
hartnn
  • hartnn
if x is position, dy/dx gives velocity and d2y/dx2 gives the acceleration. so, it does.
inkyvoyd
  • inkyvoyd
no, dx/dt gives velocity - velocity is the derivative of distance with respect to time right?
hartnn
  • hartnn
oh, yes, sorry....
inkyvoyd
  • 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 >.<
inkyvoyd
  • 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
hartnn
  • hartnn
i am sure , just realised , that d2y/dx2 is NOT acceleration
inkyvoyd
  • inkyvoyd
Okay - how should I prove to my teacher that d^2y/dx^2 is NOT acceleration?
hartnn
  • hartnn
units.....x and y are distances..... d2y/dx2 is dimensionless. it doesn't have the unit of acceleration.
inkyvoyd
  • inkyvoyd
Okay - any other ways? I tried emailing her saying units don't add up but I'm not sure if she read. >.<
hartnn
  • hartnn
d2y/dx2 gives change in dy/dx w.r.t x which is in no way, acceleration
inkyvoyd
  • inkyvoyd
because the change in y with respect to x isn't velocity right?
hartnn
  • hartnn
yeah, it isn't velocity , its just the slope
inkyvoyd
  • 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 >.<
hartnn
  • hartnn
that equality doesn't hold!
inkyvoyd
  • inkyvoyd
it does, but it has nothing to do with the definition of acceleration in two dimensions
inkyvoyd
  • 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.
hartnn
  • hartnn
you know the right thing,thats enough
inkyvoyd
  • inkyvoyd
my grade is hurt though >.<
inkyvoyd
  • inkyvoyd
Ah well I'll think of something - thank you very much!
hartnn
  • hartnn
hmm...welcome ^_^
amistre64
  • 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
hartnn
  • hartnn
product rule for y'' ?

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