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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?"
 2 years ago
 2 years ago
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?"
 2 years ago
 2 years ago

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inkyvoyd Group TitleBest ResponseYou've already chosen the best response.1
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 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.1
@hartnn , can you help me please?
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.1
I just need to know what I should be calculating for acceleration, and why. thanks!
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
i would have calculated d^2y/dx^2 because i didn't know the sqrt formula :P
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.1
But, d^2y/dx^2 is the movement of y with respect to x  it doesn't say the particle is accelerating does it?
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
if x is position, dy/dx gives velocity and d2y/dx2 gives the acceleration. so, it does.
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.1
no, dx/dt gives velocity  velocity is the derivative of distance with respect to time right?
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
oh, yes, sorry....
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.1
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 >.<
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
i am sure , just realised , that d2y/dx2 is NOT acceleration
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.1
Okay  how should I prove to my teacher that d^2y/dx^2 is NOT acceleration?
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
units.....x and y are distances..... d2y/dx2 is dimensionless. it doesn't have the unit of acceleration.
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.1
Okay  any other ways? I tried emailing her saying units don't add up but I'm not sure if she read. >.<
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
d2y/dx2 gives change in dy/dx w.r.t x which is in no way, acceleration
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.1
because the change in y with respect to x isn't velocity right?
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
yeah, it isn't velocity , its just the slope
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.1
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 >.<
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
that equality doesn't hold!
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.1
it does, but it has nothing to do with the definition of acceleration in two dimensions
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
you know the right thing,thats enough
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.1
my grade is hurt though >.<
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.1
Ah well I'll think of something  thank you very much!
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
hmm...welcome ^_^
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
product rule for y'' ?
 2 years ago
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