Question

The displacement of a body with time t is related as S=ae^6t - be^-4t. Find the rate of change of displacement with respect to time t.

Answer 1

rate of change of displacement with respect to time t. means u have to find \[\frac{ ds }{ dt }\]

Answer 2

how to ds/dt S=ae^6t - be^-4t ?

Answer 3

s is given now just differentiate it with respect to t

Answer 4

You just need to apply chain rule for exponential function. do you know that rule?

Answer 5

and (ds/dt) means the velocity of the particle

Answer 6

\[S=ae ^{6t}-be ^{-4t}\]

Answer 7

\(\Large \dfrac{d}{dt}e^t = e^t\)

Answer 8

hmm
then u can find (ds/dt) can't u :)

Answer 9

Have you been taught calculus before?

Answer 10

Yup

Answer 11

\[\huge \frac{ d }{ dt}e^{at}=ae^{at}\] as @geerky42 said

Answer 12

so what now ..!

Answer 13

now just simple differentiation using chain rule
\[\large \frac{ ds }{ dt }=\frac{ d }{ dt }[ae^{6t}-be^{-4t}]\]

Answer 14

\[ S \prime= ae ^{6t}.(6)-be ^{-4t}.(-4)\]
right?

Answer 15

yup good :)

Answer 16

what's next?

Answer 17

nothing ds/dt is evaluated :)

Answer 18

But the question says find the rate of change of displacement with respect to time t.
:|

Answer 19

yeah let s is the displacement then rate of change of displacement with respect to time t.
is \[\frac{ \Delta s }{ \Delta t }=\frac{ ds }{ dt}\] ok???

Answer 20

ammmmmm no I don't understand

Answer 21

I will come back BRB

Answer 22

let displacement of a particle at time T1 is S1 and sometime later at T2 it is S2 then change in time =(T2-T1) and change in displacement=(S2-S1)
so rate of change of displacement with respect to time is \[\frac{ S2-S1 }{ T2-T1}=\frac{ \Delta S }{ \Delta T}=ds/dt\]
do u get it now?

Answer 23

I don't think soso realizes that he has the answer already, but I'm not sure what he doesn't understand.

Answer 24

yeah mee too

Answer 25

:(