Question

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?"

Answer 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?

Answer 2

@hartnn , can you help me please?

Answer 3

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

Answer 4

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

Answer 5

But, d^2y/dx^2 is the movement of y with respect to x - it doesn't say the particle is accelerating does it?

Answer 6

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

Answer 7

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

Answer 8

oh, yes, sorry....

Answer 9

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 >.<

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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 >.<

Answer 19

that equality doesn't hold!

Answer 20

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

Answer 21

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.

Answer 22

you know the right thing,thats enough

Answer 23

my grade is hurt though >.<

Answer 24

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

Answer 25

hmm...welcome ^_^

Answer 26

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

Answer 27

product rule for y'' ?