A model car moves so that its distance, x centimetres, from a fixed point O after time t seconds is given by x=1/2t^4 - 20t^2 + 66t, , 0 </= t </= 4. i) Verify that x has a stationary value...

A model car moves so that its distance, x centimetres, from a fixed point O after time t seconds is given by x=1/2t^4 - 20t^2 + 66t,

, 0 </= t </= 4.

i) Verify that x has a stationary value when t=3, and determine whether this stationary value is a maximum value or a minimum value.

ii) Find the rate of change of x with respect to t when t=1

iii) Determine whether the distance of the car from O is increasing or decresasing at the instant when t=2

beckden | Certified Educator

x = 1/2t^4 - 20t^2 + 66t, 0<=t<=4

i) At t=3 x = 1/2(81) - 20(9) + 66(3) = 40.5 - 180 + 198 = 58.5 cm.

At t = 3.01 x = 58.500706

At t = 2.99 x = 58.500694

So x > 58 when [0, 3) and x > 58 when (3,4] and x = 58.5 at x=3 so this must be a minimum by the intermediate value theorem.

ii)  Find the rate of change of x with respect to t at t = 1.  This is the instantaneous velocity.  We can find the distance at t = 1 + h and subtract off the distance at t = 1.  If we divide this by the time between these measurements then let h go to zero we will get the rate of change.

x(1+h) = (1+h)^4/2 - 20(1+h)^2 + 66(1+h) simplifying we get

x(1+h) = 1/2(1 + 4h + 6h^2 + 4h^3 + h^4) - 20(1 + 2h + h^2) + 66 + 66h

x(1+h) = 1/2 + 2h + 3h^2 + 2h^3 + h^4/2 - 20 - 40h - 20h^2 + 66 + 66h

x(1+h) = 93/2 + 28h - 17 h^2 + 2h^3 + h^4/2

Now we find x(1)

x(1) = 1/2 - 20 + 66 = 93/2

x(1+h) - x(1) = 28h - 17 h^2 + 2h^3 + h^4/2

Now

(x(1+h) - x(1))/h = 28 - 17 h + 2h^2 + h^3/2

If we let h-> 0 we get

rate of change with respect to t = 28 cm/sec

iii) Is the cars distance from O increasing or decreasing at t = 2.

At t = 2  x = 60, at t = 1.99 x = 59.979 cm, at t = 2.01 x = 60.019 cm, so x is increasing at t = 2.

So

i) at t=3 x is a minimum.

ii) at t = 1 the rate of change of x with respect to t is 28cm/sec.

iii) x is increasing at t = 2