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