Find the unit tangent vector T(t) and find a set of parametric equations for the line tangent to the space curve at point P.
r(t)=ti+t^(2)j+tk, P(0,0,0)
T(t) = r'/|r'| = <1,2t,1>/√(4t^2+2)
P(0,0,0) = r(0)
T(0) = <1,0,1>/√2
r(0) = 0, so
the line is <0,0,0>+<1/√2,0,1/√2>t
How do you know that t=0 at (0,0,0)? How do you find it?
well, what value of t makes t and t^2 all zero?
r(t)=〈 2 sin t, 2 cos t, 4 sin^(2) t 〉 at the point (1, sqrt(3),1)
For this it is t=pi/6 to (1,sqrt(3),1). Can you show me how it is t=pi/6?
well, we want
2sint = 1, so sint = 1/2, so t=pi/6
Now it's just a matter of confirming that this also gives correct values for j and k.
It does.
To find the unit tangent vector T(t), we need to first find the derivative of the position vector r(t) with respect to t.
Differentiating each component of r(t), we have:
r'(t) = i + 2tj + k
Next, we find the magnitude of the velocity vector r'(t):
| r'(t) | = sqrt((1)^2 + (2t)^2 + (1)^2) = sqrt(1 + 4t^2 + 1) = sqrt(2(1 + 2t^2))
To find the unit tangent vector T(t), we divide the velocity vector r'(t) by its magnitude:
T(t) = r'(t) / | r'(t) | = (i + 2tj + k) / sqrt(2(1 + 2t^2))
Now, let's find the set of parametric equations for the line tangent to the space curve at point P.
The position vector of point P is r(0) = 0i + 0j + 0k = 0.
The parametric equations for a line can be given by:
x = x0 + at
y = y0 + bt
z = z0 + ct
Since the line is tangent to the curve at point P, the direction vector of the line is the same as the unit tangent vector at that point. Therefore, we have:
a = 1
b = 2t
c = 1
Substituting the values, the parametric equations for the line tangent to the space curve at point P are:
x = 0 + t
y = 0 + 2t^2
z = 0 + t
Thus, the unit tangent vector T(t) is (i + 2tj + k) / sqrt(2(1 + 2t^2)) and the parametric equations for the line tangent to the space curve at point P are x = t, y = 2t^2, and z = t.