Find parametric equations for the tangent line to the curve of intersection of the paraboloid
z = x2 + y2
and the ellipsoid
7x2 + 2y2 + 6z2 = 33
at the point
(−1, 1, 2).
(Enter your answer as a comma-separated list of equations. Let x, y, and z be in terms of t.)
Well, we all know that math can be quite the circus act! So, let's put on a show and find those parametric equations for the tangent line!
To begin, we need to find the direction vector of the tangent line. By taking the partial derivatives of both equations with respect to x, y, and z, respectively, we can find the direction vector.
For the paraboloid, we have:
dz/dx = 2x
dz/dy = 2y
For the ellipsoid, we have:
14x + 6z dx/dt = 0
4y dy/dt + 12z dz/dt = 0
Now, let's plug in the given point (-1, 1, 2) and solve the system of equations:
2(-1) dx/dt = 0
4(1) dy/dt + 12(2) dz/dt = 0
Simplifying, we find:
dx/dt = 0
dy/dt + 6 dz/dt = 0
Now, we can express x, y, and z in terms of t using the given equations:
x = -1
y = t
z = 2 - 3t
Finally, we have our parametric equations for the tangent line:
x = -1
y = t
z = 2 - 3t
So, ladies and gentlemen, the show is complete! The parametric equations for the tangent line to the curve of intersection of the paraboloid and the ellipsoid at the point (-1, 1, 2) are x = -1, y = t, and z = 2 - 3t. Enjoy the circus!
To find the parametric equations for the tangent line to the curve of intersection of the paraboloid and the ellipsoid at the given point, we need to find a vector that is tangent to both surfaces at that point.
Step 1: Find the gradient vectors of both surfaces at the given point.
The gradient vector of the paraboloid z = x^2 + y^2 is given by:
∇f = [∂f/∂x, ∂f/∂y, ∂f/∂z] = [2x, 2y, -1]
The gradient vector of the ellipsoid 7x^2 + 2y^2 + 6z^2 = 33 is given by:
∇g = [∂g/∂x, ∂g/∂y, ∂g/∂z] = [14x, 4y, 12z]
Step 2: Substitute the coordinates of the given point into the gradient vectors to find the tangent vectors.
For the paraboloid, substituting (-1, 1, 2) into the gradient vector gives:
∇f(-1, 1, 2) = [-2, 2, -1]
For the ellipsoid, substituting (-1, 1, 2) into the gradient vector gives:
∇g(-1, 1, 2) = [-14, 4, 24]
Step 3: Take the cross product of the tangent vectors to find a vector that is tangent to both surfaces at the given point.
Taking the cross product of [-2, 2, -1] and [-14, 4, 24], we get:
Tangent vector = [-52, -26, -28]
Step 4: Write the parametric equations for the tangent line.
Since the vector [-52, -26, -28] is tangent to both surfaces at the given point, we can write the parametric equations for the tangent line as:
x = -1 - 52t
y = 1 - 26t
z = 2 - 28t
Therefore, the parametric equations for the tangent line are:
x = -1 - 52t,
y = 1 - 26t,
z = 2 - 28t.
To find the parametric equations for the tangent line to the curve of intersection of the paraboloid and the ellipsoid at the given point, we need to find the direction vector of the tangent line at that point.
First, let's find the equation of the curve of intersection by setting the equation of the paraboloid equal to the equation of the ellipsoid:
x^2 + y^2 = 7x^2 + 2y^2 + 6z^2 - 33
Simplifying this equation gives:
6x^2 + y^2 - 6z^2 = -33
Now, to find the point of intersection, substitute the given coordinates (-1, 1, 2) into the equation above:
6(-1)^2 + (1)^2 - 6(2)^2 = -33
6 + 1 - 24 = -33
-17 = -33
The equation is true, so the given point (-1, 1, 2) lies on the curve of intersection.
Next, we need to find the partial derivatives of the equations of the paraboloid and the ellipsoid with respect to x, y, and z. The resulting gradients will give us the normal vectors to the surfaces at the point of intersection:
For the paraboloid:
∇(x^2 + y^2) = (2x, 2y, 0)
For the ellipsoid:
∇(7x^2 + 2y^2 + 6z^2 - 33) = (14x, 4y, 12z)
Now, evaluate the gradients at the point (-1, 1, 2):
∇(x^2 + y^2) = (2(-1), 2(1), 0) = (-2, 2, 0)
∇(7x^2 + 2y^2 + 6z^2 - 33) = (14(-1), 4(1), 12(2)) = (-14, 4, 24)
The vector (-2, 2, 0) is normal to the paraboloid at the point of intersection, and the vector (-14, 4, 24) is normal to the ellipsoid at the point of intersection.
Since the tangent line to the curve of intersection is perpendicular to both normal vectors, the direction vector of the tangent line is given by their cross product:
d = (-2, 2, 0) x (-14, 4, 24)
Calculating the cross product:
d = (0, 48, 34) - (0, 0, -28) = (0, 48, 62)
Therefore, the direction vector of the tangent line is (0, 48, 62).
Finally, we can express the tangent line in parametric form using the point of intersection (-1, 1, 2) and the direction vector (0, 48, 62):
x = -1 + 0t
y = 1 + 48t
z = 2 + 62t
So, the parametric equations for the tangent line to the curve of intersection of the paraboloid and the ellipsoid at the point (-1, 1, 2) are:
x = -1
y = 1 + 48t
z = 2 + 62t
(a) study your text, or
(b) google for examples, such as
http://www.math.ucla.edu/~ronmiech/Calculus_Problems/32A/chap12/section6/812d55/812_55.html
and follow the same steps.