Find parametric equations for the tangent line to the curve with the given parametric equation s at the specified point.
x=t^(2)+1, y=4sqrt(t), z=e^(t^(2)-t);(2,4,1)

1 Answers

Best Answer

Frist we need to find the tangent vector r'(t)

We have

r(t)=t^(2)+1i+4sqrt(t)j+e^(t^(2)-t)k

Differentiate to get the tangent vector

r'(t)=2ti+(2)/(sqrt(t))j+(2t-1)e^(t^(2)-t)k

Note that the point (2,4,1) corresponds to t=1

Since as we put t=1 in r(t), we get (2,4,1) as the position.

r'(1)=2*1i+(2)/(sqrt(1))j+(2*1-1)e^(1^(2)-1)k

r'(1)=2i+2j+k

Remember that: Equation of a line passing through a point with position vector a, and parallel to the vector b is

r(t)=a+t b

Hence equation of the tangent line at (2,4,1) is

r(t)=langle 2,4,1 rangle + tlangle 2,2,1 rangle

r(t)=langle 2+2t,4 + 2t, 1+t rangle

Hence the parametric equations of the line are

x=2+2t, y=4+2t, z=1+t

Result:

The required equation for tangent vector is x=2+2t, y=4+2t, z=1+t.

We have

r(t)=t^(2)+1i+4sqrt(t)j+e^(t^(2)-t)k

Differentiate to get the tangent vector

r'(t)=2ti+(2)/(sqrt(t))j+(2t-1)e^(t^(2)-t)k

Note that the point (2,4,1) corresponds to t=1

Since as we put t=1 in r(t), we get (2,4,1) as the position.

r'(1)=2*1i+(2)/(sqrt(1))j+(2*1-1)e^(1^(2)-1)k

r'(1)=2i+2j+k

Remember that: Equation of a line passing through a point with position vector a, and parallel to the vector b is

r(t)=a+t b

Hence equation of the tangent line at (2,4,1) is

r(t)=langle 2,4,1 rangle + tlangle 2,2,1 rangle

r(t)=langle 2+2t,4 + 2t, 1+t rangle

Hence the parametric equations of the line are

x=2+2t, y=4+2t, z=1+t

Result:

The required equation for tangent vector is x=2+2t, y=4+2t, z=1+t.