Problem 34 - The Calculus with Analytic Geometry

Question

Find equations of the tangent line to the curve of intersection of the surface \(x^{2}+y^{2}+z^{2}=9\) with the plane \(y=2\) at the point \((1,2,2)\).

Step-by-Step Solution

Answer

The parametric equations of the tangent line are: x = 1 - 4t, y = 2, z = 2 + 2t.

Step 1: Identify the surface equation

The surface equation given is x^{2}+y^{2}+z^{2}=9 . We need to find the curve of intersection with the given plane.

Step 2: Substitute y in the surface equation

Substitute the value y = 2 into the surface equation to get x^{2}+2^{2}+z^{2}=9.

Step 3: Simplify the equation

Simplify the equation to get x^{2} + z^{2} = 5.

Step 4: Find the gradient of the surface

To find the tangent line we need the normal vector to the surface at the point (1,2,2). The gradient of the surface ∇(x^{2}+y^{2}+z^{2}) = (2x, 2y, 2z). At the point (1,2,2), the gradient is (2*1, 2*2, 2*2) = (2, 4, 4).

Step 5: Find direction vector for the curve

Find the direction vector to the curve of intersection by taking the cross product of the gradient of the surface (2, 4, 4) and the normal vector to the plane y = 2 which is (0, 1, 0).

Step 6: Compute the cross product

Compute the cross product (2, 4, 4) x (0, 1, 0) = (4*0 - 4*1, 4*0 - 2*0, 2*1 - 4*0) = (-4, 0, 2). Thus, the direction vector is (-4, 0, 2).

Step 7: Write parametric equations of the tangent line

Use the point (1, 2, 2) and direction vector (-4, 0, 2) to write the parametric equations of the tangent line: x = 1 - 4t, y = 2, z = 2 + 2t.

Key Concepts

Intersection of Surfaces

The intersection of surfaces is where two different geometric shapes, represented by their equations, meet in 3-dimensional space.
To find the intersection of the surface defined by \(x^2 + y^2 + z^2 = 9\) with the plane \(y = 2\), we substitute \(y = 2\) into the surface equation.
This reduces our problem to solving \(x^2 + 2^2 + z^2 = 9\), which simplifies to \(x^2 + z^2 = 5\).
The resulting equation represents a circle in the \(xz\) plane, showing the curve along which the two surfaces intersect.

Gradient Vector

The gradient vector is crucial when working with surfaces in 3D.
It points in the direction of the steepest increase of a function and is perpendicular to level surfaces.
To find the tangent line, we need the gradient of our surface at the point of interest. For the surface \(x^2 + y^2 + z^2 = 9\), the gradient vector is ∇\(f\) = (2x, 2y, 2z).
At the point (1, 2, 2), it becomes (2*1, 2*2, 2*2) = (2, 4, 4). This vector is normal to our surface at (1, 2, 2).

Cross Product

The cross product of two vectors gives a vector perpendicular to the plane containing them.
To find the direction vector of the tangent line, we take the cross product of the surface's gradient vector (2, 4, 4) and the normal vector to the plane y = 2, which is (0, 1, 0).
Computing the cross product: \((2, 4, 4) \times (0, 1, 0)= (-4, 0, 2)\).
This direction vector is along the tangent to the curve of intersection at point (1, 2, 2).

Parametric Equations

Parametric equations express a curve via a parameter, usually \(t\).
To write parametric equations for our tangent line, use the point (1, 2, 2) and direction vector (-4, 0, 2).
The equations describe how each coordinate changes with \(t\):

• x = 1 - 4t
• y = 2
• z = 2 + 2t

These form the parametric representation of the tangent line to the curve of intersection at (1, 2, 2).