Let `C:r(t)=x(t)hati+y(t)hatj+z(t)hatk` be a differentiable curve, i.e., `lim_(xto0) (r(t+H)-r(h))/(h)` exist for all t,
`therefore r'(t)=x'(t)hati+y'(t)hatj+z'(t)hatk`
Iff `r'(t)`, is tangent to the curve C at the point `P[x(t),y(t),z(t)] and r'(t)` points in the direction of increasing t.
Q. The point P on the curve `r(t)=(1-2t)hati+t^(2)hatj+2e^(2(t-1))hatk` at which the tangent vector `r'(t)` is parallel to the radius vector r(t) is

Options

A. `(-1,1,2)`
B. `(1,-1,2)`
C. `(-1,1,-2)`
D. `(1,1,2)`

Answer

A

Explanation

`r'(t)=-2hati+2thatj+4e^(2(t-1))hatk`
since, `r'(t)` is parallel to r(t),
so `r(t)=alphar'(t)`
`1-2t=-2alpha,t^(2)=alphat,2e^(2(t-1))`
`=4alphae^(2(t-1)),alpha=(1)/(2)`
the only value of t which satisfies all three equations is t=1.
so, r(1) is the required point (-1,1,2).