Question

Could you help me solve this problem:
A particle is moving in a line according to the equation s^2 = a + bt^2, s > 0 where a and b are positive constants. Prove that the measure of the acceleration of the particle is inversely proportional to s^3 for any t.

Answer 1

first find the first derivative of s with respect to t (i.e. find \(\frac{ds}{dt}\)) - what do you get?

Answer 2

s' = 1/2(a + bt^2)^-1/2(2bt) ?

Answer 3

looks correct but you can simplify it further.

do you know how to do implicit derivatives? If you do, then it will be easier to use that method to find \(\frac{ds}{dt}\)

Answer 4

I don't -_-

Answer 5

ok - no problem, then first try and simplify the expression you just got for \(\frac{ds}{dt}\)

Answer 6

s' = bt(a+bt^2)^-1/2 ?

Answer 7

ok, so you have correctly got to:\[\frac{ds}{dt}=\frac{bt}{\sqrt{a+bt^2}}\]now differentiate this once more to get \(\frac{d^2s}{dt^2}\)

Answer 8

Am I supposed to treat b and t as constants?

Answer 9

your question states that a and b are constants. t is the variable quantity that you are differentiating with respect to.

Answer 10

Sorry, I mean a and b

Answer 11

yes

Answer 12

I think I'm doing it wrong, can't find the answer

Answer 13

please show your steps - I can then spot where you may have made a mistake.
it might be easier for you to use the drawing tool below to write out your steps as you don't seem to know how to type equations in \(LaTeX\)

Answer 14

or upload a pic of your written attempt

Answer 15

\[s" =\frac{ (a+bt)^{-\frac{ 1 }{ 2 }}(0) - (bt)(-\frac{ 1 }{ 2 }(a+bt)^{-\frac{ 3 }{ 2 }}(0) }{ (a+bt)^{\frac{ 1 }{ 4 }} }\]

Is that correct?

Answer 16

why is the first term in the numerator multiplied by 0?
same question for 2nd term in numerator

Answer 17

you got to:\[s'=bt(a+bt^2)^{-1/2}\]so if we differentiate this we get:\[s''=(a+bt^2)^{-1/2}\times\frac{d}{dt}\left(bt\right)+bt\times\frac{d}{dt}\left((a+bt^2)^{-1/2}\right)\]agreed so far?

Answer 18

this is using the rule that if\[y=f(t)g(t)\]then:\[\frac{dy}{dt}=g(t)\times f'(t)+f(t)\times g'(t)\]

Answer 19

That's why I don't get it, I don't know that rule

Answer 20

So what to do after that?

Answer 21

you must know that rule in order to answer this question. if you haven't been taught that then I do not know what method they expect you to use.

Answer 22

look at Product Rule on this page to learn about it: http://www.mathsisfun.com/calculus/derivatives-rules.html

then try to solve again and ping me if you need further help.