3 years ago
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
Brandonellerbee26 Staff answered 3 years ago
Brandonellerbee26 Staff answered 3 years ago
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.