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Let C:vecr(t)=x(t)hati+y(t)hatj+z(t)hatk...

Let `C:vecr(t)=x(t)hati+y(t)hatj+z(t)hatk` be a differentiable curve i.e., `exists lim_(hto0) (vecr(t+h)-vecr(t))/(h) forall t`
`therefore vecr'(t)=x'(t)hati+y'(t)hatj+z'(t)hatk`
`vecr'(t)` is tangent to the curve `C` at the point `P[x(t),y(t),z(t)], vecr'(t)` points in the direction of increasing `t`.
The point `P` on the curve `vecr(t)=(1-2t)hati+t^(2)hatj+2e^(2(t-1))hatk` at which the tangent vector `vecr'(t)` is parallel to the radius vector `vecr(t)` is:

A

(a) `(-1,1,2)`

B

(b) `(1,-1,2)`

C

(c) `(-1,1,-2)`

D

(d) `(1,1,2)`

Text Solution

Verified by Experts

The correct Answer is:
A
`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).
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