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inkyvoyd
 4 years ago
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?"
inkyvoyd
 4 years ago
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?"

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inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.1I 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
 4 years ago
Best ResponseYou've already chosen the best response.1@hartnn , can you help me please?

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

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

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

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

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

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.1I 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
 4 years ago
Best ResponseYou've already chosen the best response.1I'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
 4 years ago
Best ResponseYou've already chosen the best response.1i am sure , just realised , that d2y/dx2 is NOT acceleration

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

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

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

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

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

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

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.1So 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
 4 years ago
Best ResponseYou've already chosen the best response.1that equality doesn't hold!

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

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.1yet 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
 4 years ago
Best ResponseYou've already chosen the best response.1you know the right thing,thats enough

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

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

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0spose 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
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