find equations for the osculating normal and rectifying planes to the curve x=3t-t^2, y=3t^2, z=3t+t^3 at point where t=1
standard differential geometry stuff. You surely have formulas for this in your text. Any google search will provide many discussions of the topic.
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Yes
To find the equations for the osculating normal and rectifying planes at the point where t = 1, we first need to find the position vector and the velocity vector of the curve at that point.
Given the parametric equations:
x = 3t - t^2
y = 3t^2
z = 3t + t^3
Let's start by finding the position vector r(t) and the velocity vector v(t).
1. Position vector (r(t)):
The position vector is denoted by r(t) = [x(t), y(t), z(t)].
Plugging in t = 1 into the given equations, we get:
x(1) = 3(1) - (1)^2 = 3 - 1 = 2
y(1) = 3(1)^2 = 3(1) = 3
z(1) = 3(1) + (1)^3 = 3 + 1 = 4
Therefore, the position vector at t = 1 is r(1) = [2, 3, 4].
2. Velocity vector (v(t)):
The velocity vector is the derivative of the position vector with respect to t.
v(t) = [dx/dt, dy/dt, dz/dt]
Differentiating each component of the position vector, we get:
dx/dt = 3 - 2t
dy/dt = 6t
dz/dt = 3 + 3t^2
Plugging in t = 1 into the derivatives, we get:
dx/dt = 3 - 2(1) = 3 - 2 = 1
dy/dt = 6(1) = 6
dz/dt = 3 + 3(1)^2 = 3 + 3 = 6
Therefore, the velocity vector at t = 1 is v(1) = [1, 6, 6].
Now, let's proceed to find the osculating normal plane.
3. Osculating normal plane:
The osculating normal vector is obtained by taking the derivative of the velocity vector with respect to t.
n(t) = [d^2x/dt^2, d^2y/dt^2, d^2z/dt^2]
Differentiating each component of the velocity vector, we get:
d^2x/dt^2 = -2
d^2y/dt^2 = 6
d^2z/dt^2 = 6t
Plugging in t = 1 into the second derivatives, we get:
d^2x/dt^2 = -2
d^2y/dt^2 = 6
d^2z/dt^2 = 6(1) = 6
Therefore, the osculating normal vector at t = 1 is n(1) = [-2, 6, 6].
The equation of the osculating normal plane can be written as:
-2(x - 2) + 6(y - 3) + 6(z - 4) = 0
Simplifying, we get:
-2x + 4 + 6y - 18 + 6z - 24 = 0
-2x + 6y + 6z - 38 = 0
Thus, the equation of the osculating normal plane is -2x + 6y + 6z - 38 = 0.
Next, let's find the rectifying plane.
4. Rectifying plane:
The rectifying plane contains the velocity vector.
The equation of the rectifying plane can be written as:
(v(1) · r) - (v(1) · r(1)) = 0
where · denotes the dot product.
Plugging in the known values, we get:
(1, 6, 6) · (x - 2, y - 3, z - 4) - (1, 6, 6) · (2, 3, 4) = 0
(1)(x - 2) + (6)(y - 3) + (6)(z - 4) - (1)(2) - (6)(3) - (6)(4) = 0
x - 2 + 6y - 18 + 6z - 24 - 2 - 18 - 24 = 0
x + 6y + 6z - 74 = 0
Thus, the equation of the rectifying plane is x + 6y + 6z - 74 = 0.
To summarize:
- The equation of the osculating normal plane is -2x + 6y + 6z - 38 = 0.
- The equation of the rectifying plane is x + 6y + 6z - 74 = 0.
To find the equations for the osculating normal and rectifying planes to a curve at a specific point, we need to follow these steps:
Step 1: Find the position vector of the curve at the given point.
Step 2: Find the unit tangent vector to the curve at the given point.
Step 3: Find the derivative of the unit tangent vector with respect to t (the parameter of the curve) and evaluate it at the given point to obtain the curvature vector.
Step 4: Find the cross product of the unit tangent vector and the curvature vector to obtain the osculating normal vector.
Step 5: Find the cross product of the unit tangent vector and the osculating normal vector to obtain the rectifying vector. This vector lies in the rectifying plane.
Step 6: Use the position vector and the osculating normal vector to find the equation of the osculating normal plane.
Step 7: Use the position vector and the rectifying vector to find the equation of the rectifying plane.
Now let's apply these steps to the given curve x = 3t - t^2, y = 3t^2, z = 3t + t^3 and find the equations for the osculating normal and rectifying planes at the point where t = 1.
Step 1: Find the position vector of the curve at the given point:
At t = 1, the position vector of the curve is r(1) = (2, 3, 4).
Step 2: Find the unit tangent vector to the curve at the given point:
The unit tangent vector is given by T(t) = (dx/dt, dy/dt, dz/dt) / |(dx/dt, dy/dt, dz/dt)|.
Calculating the derivatives, we have:
dx/dt = 3 - 2t, dy/dt = 6t, dz/dt = 3 + 3t^2.
Substituting t = 1, we get:
dx/dt = 1, dy/dt = 6, dz/dt = 6.
The magnitude of the tangent vector is |T(1)| = sqrt(1^2 + 6^2 + 6^2) = sqrt(1 + 36 + 36) = sqrt(73).
Therefore, the unit tangent vector at t = 1 is T(1) = (1/sqrt(73), 6/sqrt(73), 6/sqrt(73)) = (1/√73, 6/√73, 6/√73).
Step 3: Find the derivative of the unit tangent vector with respect to t and evaluate it at t = 1 to obtain the curvature vector:
The curvature vector is given by K(t) = dT(t)/dt.
Differentiating T(t) with respect to t, we have:
d/dt(T(t)) = (d^2x/dt^2, d^2y/dt^2, d^2z/dt^2) / |(dx/dt, dy/dt, dz/dt)| - ((dx/dt, dy/dt, dz/dt) ⋅ d/dt(dx/dt, dy/dt, dz/dt)) / |(dx/dt, dy/dt, dz/dt)|^3.
Calculating the derivatives, we have:
d^2x/dt^2 = -2, d^2y/dt^2 = 6, d^2z/dt^2 = 6t.
Substituting t = 1, we get:
d^2x/dt^2 = -2, d^2y/dt^2 = 6, d^2z/dt^2 = 6.
Substituting these values into the formula for K(t), we have:
K(1) = (-2/sqrt(73), 6/sqrt(73), 6/sqrt(73)).
Step 4: Find the cross product of the unit tangent vector and the curvature vector to obtain the osculating normal vector:
The osculating normal vector is given by N(t) = T(t) × K(t).
Substituting the values we found earlier, we have:
N(1) = (1/√73, 6/√73, 6/√73) × (-2/√73, 6/√73, 6/√73).
Calculating the cross product, we get:
N(1) = (0, -2/√73, 2/√73).
Step 5: Find the cross product of the unit tangent vector and the osculating normal vector to obtain the rectifying vector:
The rectifying vector is given by R(t) = T(t) × N(t).
Using the values we calculated earlier, we have:
R(1) = (1/√73, 6/√73, 6/√73) × (0, -2/√73, 2/√73).
Calculating the cross product, we get:
R(1) = (-8/73, 0, -8/73).
Step 6: Use the position vector and the osculating normal vector to find the equation of the osculating normal plane:
The equation of a plane passing through a point (x0, y0, z0) with normal vector (a, b, c) is ax + by + cz = d, where d is calculated by substituting the point into the equation.
Using the position vector r(1) = (2, 3, 4) and the osculating normal vector N(1) = (0, -2/√73, 2/√73), we can obtain the equation of the osculating normal plane:
0x - (2/√73)y + (2/√73)z = d.
Substituting the coordinates of the point (2, 3, 4) into the equation, we get:
-(2/√73)(3) + (2/√73)(4) = d.
Simplifying, we find:
d = 2/√73.
Therefore, the equation of the osculating normal plane is:
-(2/√73)y + (2/√73)z = 2/√73.
Step 7: Use the position vector and the rectifying vector to find the equation of the rectifying plane:
Using the position vector r(1) = (2, 3, 4) and the rectifying vector R(1) = (-8/73, 0, -8/73), we can obtain the equation of the rectifying plane:
(-8/73)x + (-8/73)z = d.
Substituting the coordinates of the point (2, 3, 4) into the equation, we get:
(-8/73)(2) + (-8/73)(4) = d.
Simplifying, we find:
d = -16/73.
Therefore, the equation of the rectifying plane is:
(-8/73)x + (-8/73)z = -16/73.
So, the equations for the osculating normal and rectifying planes to the curve x = 3t - t^2, y = 3t^2, z = 3t + t^3 at the point where t = 1 are:
Osculating Normal Plane: -(2/√73)y + (2/√73)z = 2/√73.
Rectifying Plane: (-8/73)x + (-8/73)z = -16/73.