Minimum Speed given Position Vector

Question

The position function of a particle is r(t) = <-t^2 , 5t, t^2 -4t>.

At what time is the speed minimum.

I took the derivative of the position function, then took the magnitude of that and set it equal to 0 to find the minimum speed... but when I tried to do this, I couldn't isolate the t value.

If you could explain how you get to your answer, that would be helpful!

Answer 1

r(t) = <-t^2 , 5t, t^2 -4t>

r'(t) = <-2t , 5, 2t -4>

v = |r'(t)| = √[(-2t)^2+(5)^2+(2t-4)^2] = √(4t^2+25+4t^2-16t+16) = √(8t^2-16t+41)

v' = 1/ [2√(8t^2-16t+41)] (16t-16) <-chain rule

v'(t) = 0 <=> 16t-16 = 0 since 8t^2-16t+41 >0 for all real t

∴ t= 1 is a turning point. Max or min is to be determined below.

Proof of Minimum: (We use first derivatives test here but not the second derivatives test to avoid applying quotient rule on those complicated expressions)

v'(1^-)= 1/ [2√(8(1^-)^2-16(1^-)+41)] (16(1^-)-16) <0 , v(t) is decreasing near 1^-

v'(1⁺)= 1/ [2√(8(1⁺)^2-16(1⁺)+41)] (16(1⁺)-16) >0, v(t) is increasing near 1⁺

Since the v(t) goes from decreasing to increasing at t=1, then that point is a local minimum.

Answer 2

isn't v(t)=dr/dt=(-2t,5,2t-4)

|v(t)|²=4t²+25+4t²-16t+16

is there a minimum for |v(t)|²?

d(8t²-16t+41)/dt=16t-16,,,,critical point at t=1

d²(8t²-16t+41)/dt²=16 second derivative =16>0 for all x

critical point is a minimum