Find a vector equation for the tangent line to the curve of intersection of the cylinders x^2 + y^2 = 25 and y^2 + z^2 = 20 at the point (3, 4, 2).
eliminating y^2, we have
x^2 + 20-z^2 = 25
x^2 - z^2 = 5
Now just find the tangent line in the usual manner.
To find the vector equation for the tangent line to the curve of intersection between the cylinders, we first need to find the equations of the curves themselves.
The equation of the first cylinder is x^2 + y^2 = 25, which represents a circular cross-section in the xy-plane with a radius of 5 units and its center at the origin (0, 0, 0).
The equation of the second cylinder is y^2 + z^2 = 20, which represents a circular cross-section in the yz-plane with a radius of √20 units and its center at the origin (0, 0, 0).
To find the curve of intersection between the two cylinders, we need to solve these equations simultaneously. We can do this by substituting one equation into the other.
Substituting y^2 = 25 - x^2 from the first equation into the second equation, we get:
(25 - x^2) + z^2 = 20
z^2 = 20 - 25 + x^2
z^2 = -5 + x^2
Now we have the equation for the curve of intersection in terms of x and z.
To find the tangent line at the point (3, 4, 2), we need to calculate the gradient vector of the curve at that point.
Let's differentiate the equation z^2 = -5 + x^2 with respect to x:
2z * dz/dx = 2x
dz/dx = x/z
At the point (3, 4, 2), x = 3 and z = 2. Plugging these values into the equation, we get:
dz/dx = 3/2
So the slope of the tangent line at the point (3, 4, 2) is 3/2.
To find the direction vector of the tangent line, we need to use the coefficients of x, y, and z in the equation of the tangent plane.
Let's write the equation of the tangent plane in the form Ax + By + Cz = D, where A, B, C, and D are constants.
At the point (3, 4, 2), we substitute the values into the equation:
A * 3 + B * 4 + C * 2 = D
Since the curve lies on both cylinders, it also satisfies both equations:
(3)^2 + (4)^2 = 25
(4)^2 + (2)^2 = 20
9 + 16 = 25
16 + 4 = 20
These equations simplify to:
25 = 25
20 = 20
This tells us that the point (3, 4, 2) lies on the curves of both cylinders and satisfies the equation of the plane.
Now we can obtain the coefficients A, B, C, and D by differentiating the equations of the cylinders and substituting the coordinates (3, 4, 2) into the resulting equations.
The partial derivatives with respect to x, y, and z are as follows:
∂(x^2 + y^2)/∂x = 2x
∂(x^2 + y^2)/∂y = 2y
∂(y^2 + z^2)/∂y = 2y
∂(y^2 + z^2)/∂z = 2z
At the point (3, 4, 2), we substitute the values into these equations:
∂(x^2 + y^2)/∂x = 2 * 3 = 6
∂(x^2 + y^2)/∂y = 2 * 4 = 8
∂(y^2 + z^2)/∂y = 2 * 4 = 8
∂(y^2 + z^2)/∂z = 2 * 2 = 4
We can then substitute these values into the equation of the tangent plane:
A * 3 + B * 4 + C * 2 = D
6A + 8B + 8C = D
Now we have the coefficients A, B, C, and D for the equation of the tangent plane.
Finally, the vector equation for the tangent line is given by:
r(t) = (3, 4, 2) + t * (A, B, C)
where (A, B, C) is the direction vector of the tangent line, and t is a parameter that represents any point on the line.