A point moves on d x-axis in such a way that its velocity at time t (t>0) is given by v=(lnt/t)...?

Question

At what value of t does v attain its maximim?
A) 1
B) e^½
C) e
D) e^(3/2)
E) there's no maximum value for v

Answer 1

(c)

take the derivative, set it = 0

Answer 2

Velocity is maxi-um when the derivative changes from positive to negative.

I don't have a pencil on me but find the derivative of ln/t and see where the sign changes from positive to negative (by graphing).

Answer 3

Classic differentiation question!
Solution:-
given, velocity v = (ln(t)/t)
take first derivative w.r.t. time t on both sides
thus dv/dt = (d(ln(t))/dt)*(1/t) + ln(t)*d(1/t)/dt
or dv/dt = 1/t^2 -ln(t)/t^2
for maximum velocity, dv/dt must be equal to 0
Hence dv/dt = 0 => 1/t^2 - ln(t)/t^2 = 0 => 1 - ln(t) = 0
=> ln(t) =1 => t = e
Hence, the correct answer is C) t = e

Answer 4

v´= 1/t^2 *( 1 -ln t)
v´=0 if ln t=1 so t=e
the sign of v´is +++++++e----------- so at t=e we have a relative maximum which is v=1/e

Answer 5

thankyou for all the answers.