Home
Class 12
MATHS
Let C:r(t)=x(t)hati+y(t)hatj+z(t)hatk be...

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 tangent vector to `r(t)=2t^(2)hati+(1-t)hatj+(3t^(2)+2)hatk` at (2,0,5) is

A

`4hati+hatj-6hatk`

B

`4hati-hatj+6hatk`

C

`2hati-hatj+6hatk`

D

`2hati+hatj-6hatk`

Text Solution

Verified by Experts

The correct Answer is:
B
(2,0,5) corresponding to r(1) and r'9t)=`4thati-thatj+6t hatk`
so, the required tangent vector is `r'(1)=4hati-hatj+6hatk`.
Promotional Banner

Topper's Solved these Questions

  • VECTOR ALGEBRA

    ARIHANT MATHS|Exercise Exercise (Matching Type Questions)|1 Videos

  • VECTOR ALGEBRA

    ARIHANT MATHS|Exercise Exercise (Single Integer Answer Type Questions)|7 Videos

  • VECTOR ALGEBRA

    ARIHANT MATHS|Exercise Exercise (Statement I And Ii Type Questions)|3 Videos

  • TRIGONOMETRIC FUNCTIONS AND IDENTITIES

    ARIHANT MATHS|Exercise Exercise (Questions Asked In Previous 13 Years Exam)|19 Videos

Similar Questions

Explore conceptually related problems

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

Find the locus of the point (t^(2)-t+1,t^(2)+t+1),t in R

Find the Cartesian from the equation of the plane vecr=(s-2t)hati+(3-t)hatj+(2s+t)hatk .

A unit tangent vector at t=2 on the curve x=t^(2)+2, y=4t-5 and z=2t^(2)-6t is

The slope of the tangent to the curve x =1/t, y =t - t/t , at t=2 is ………….

Write the equation of a tangent to the curve x=t, y=t^2 and z=t^3 at its point M(1, 1, 1): (t=1) .

The slope of the tangent to the curve x=t^(2)+3t-8,y=2t^(2) -2t -5 at the point (2, -1), is