If (2, 3, 5) is one end of a diameter of the sphere `x^2+""y^2+""z^2-6x-12 y-2z""+""20""=""0` , then the coordinates of the other end of the diameter are (1) `(4,""9,-3)` (2) `(4,-3,""3)` (3) `(4,""3,""5)` (4) `(4,""3,-3)`

Options

A. `x(t) gt 0` for all `t gt 0`
B. `v(t) gt 0` for all `t gt 0`
C. `a(t) gt 0` for all `t gt 0`
D. `v(t)` lies between `0` and `2`

Answer

A, D

Explanation

Given, `x = t - sint`
velocity `v = (dx)/(dt) = (d)/(dt) [t - sin t]`
`= 1- "cost"`
Acceleration `a = (dv)/(dt) = (d)/(dt) [1-"cost"] = sint`
As acceleration `a gt 0` for all `t gt 0`
Hence, `x(t) gt 0` for all `t gt 0`
Velocity `v = 1 - cos t`
when, `cos t = 1, "velocity" v = 0`
`v_(max) = 1 - (cos t)_(min) = 1 - (-1) = 2`
`v_(min) = 1 - (cos t)_(max) = 1 - 1 = 0`
Hence, v lies between 0 and 2
Acceleration `a = (dv)/(dt) = - sint`
When , `t = 0, x = 0 , x = +1 , a = 0`
When, `t = (pi)/(2), x = 1, v = 0, a = - 1`
When `t = pi, x = 0, x = -1, a = 1`
When `t = 2pi, x = 0, x = 0, a = 0`