Question

Given v(s)=ds/dt=2s^2+3. Find a=d^2s/dt^2 as a function of s.

Answer 1

4s?

Answer 2

hmm ya.

Answer 3

so what if the question said: given v(t)=ds/dt=2t^2+3. Find a=d^2s/dt^2 as a function of t.

Answer 4

i just want to clarify\[\frac{ds}{dt}=2s^2+3\]\[\frac{d^2s}{dt^2}=?\]

Answer 5

yes

Answer 6

the thing that is throwing me off is the "as a function of s" or "as a function of t" part

Answer 7

According to what you did, shouldn't the [2s ^{2}+3\] be squared?

Answer 8

2s^2+3

Answer 9

ahh wait its not clear

Answer 10

i think u made a small error here:
\[a=\frac{d^2s}{dt^2}=\frac{d}{dt}(\frac{ds}{dt})=\frac{d}{ds}(\frac{ds}{dt})\frac{dt}{ds}\]

Answer 11

\[(2s ^{2}+3)^{2}\]

Answer 12

\[a=[\frac{d}{ds}(\frac{ds}{dt})]\times\frac{ds}{dt}=[\frac{d}{ds}(2s^2+3)]\times (2s^2+3)=4s(2s^2+3)\]

Answer 13

d/ds=4s, right?

Answer 14

right

Answer 15

so 4s(2s^2+3)x(2s^2+3) equals what?

Answer 16

shouldn't it equal 4s(2s^2+3)^2

Answer 17

using chain rule\[a=\frac{d^2s}{dt^2}=\frac{d}{dt}(\frac{ds}{dt})=[\frac{d}{ds}(\frac{ds}{dt})]\times[\frac{ds}{dt}]\]

Answer 18

\[a=[\frac{d}{ds}(\frac{ds}{dt})]\times[\frac{ds}{dt}]=[\frac{d}{ds}(2s^2+3)]\times(2s^2+3)\] \(=[2s^2+3]'(2s^2+3)=4s(2s^2+3)\)

Answer 19

ohhhhhhhhhhhhhhhhhhhhhhh

Answer 20

sorry dude, totally was reading it wrong

Answer 21

np :)

Answer 22

thank you