Question

Line tangent to an intersection curve Consider the paraboloid \(z=x^{2}+3 y^{2}\) and the plane \(z=x+y+4,\) which intersects the paraboloid in a curve \(C\) at (2,1,7) (see figure). Find the equation of the line tangent to \(C\) at the point \((2,1,7) .\) Proceed as follows. a. Find a vector normal to the plane at (2,1,7) b. Find a vector normal to the plane tangent to the paraboloid at (2,1,7) c. Argue that the line tangent to \(C\) at (2,1,7) is orthogonal to both normal vectors found in parts (a) and (b). Use this fact to find a direction vector for the tangent line. d. Knowing a point on the tangent line and the direction of the tangent line, write an equation of the tangent line in parametric form.

Short answer

Question: Find the parametric equations for the line tangent to the curve C formed by the intersection of the paraboloid \(z=x^2+3y^2\) and the plane \(z=x+y+4\) at the point (2,1,7). Answer: The parametric equations for the line tangent to C at (2,1,7) are \(x(t)=2+ct\), \(y(t)=1-ct\), and \(z(t)=7+ct\), where \(\vec{v}=\langle c,-c,c\rangle\) is the direction vector of the tangent line.

Step-by-step solution

1. Find the normal vector to the plane at (2,1,7)

The given plane has the equation \(z=x+y+4\). We can rewrite this equation as \(x+y-z+4=0\). Now, we can find the normal vector by looking at the coefficients of x, y, and z, which are (1,1,-1). So, the normal vector to the plane is \(\langle 1,1,-1\rangle\).

2. Find the normal vector to the tangent plane to the paraboloid at (2,1,7)

The given paraboloid has the equation \(z=x^2+3y^2\). We can find the gradient of this surface as \(\nabla f = \langle 2x, 6y, -1\rangle\). At the point (2,1,7), the gradient of the paraboloid is \(\langle 4,6,-1\rangle\). This gradient vector is normal to the tangent plane of the paraboloid at the given point.

3. Orthogonality of the tangent line and normal vectors

The tangent line to the curve C at the point (2,1,7) must be orthogonal to the normal vectors found in parts (a) and (b), as it lies within both the plane and the tangent plane to the paraboloid. Let's denote the direction vector of the tangent line as \(\vec{v}=\langle a,b,c\rangle\). Then, in order for it to be orthogonal to both normal vectors, the following must be true: \(\vec{v} \cdot \langle 1,1,-1\rangle = 0\) \(\vec{v} \cdot \langle 4,6,-1\rangle = 0\)

4. Find a direction vector for the tangent line

Using the two orthogonality equations above, we get: \(a + b - c = 0\) \(4a + 6b - c = 0\) Solving this system of linear equations, and expressing in terms of the free variable c, we get: \(a = c, b = -c, c = c\) The direction vector can be written as \(\vec{v}=\langle a,b,c\rangle = \langle c,-c,c\rangle\).

5. Write the tangent line equation in parametric form

Now that we know the point on the tangent line (2,1,7) and its direction vector \(\vec{v}=\langle c,-c,c\rangle\), we can write the equation of the tangent line in parametric form as follows: \(x(t) = 2 + ct\) \(y(t) = 1 - ct\) \(z(t) = 7 + ct\) Thus, the parametric equations for the line tangent to C at (2,1,7) are: \(x(t)=2+ct\), \(y(t)=1-ct\), and \(z(t)=7+ct\).

Key concepts

Paraboloid

A paraboloid is a three-dimensional surface which can be thought of as a generalization of a parabola. It is defined by a quadratic equation in two variables, typically expressed as \(z = ax^2 + by^2\). This specific shape can be either an elliptic paraboloid, which opens upward like a dish, or a hyperbolic paraboloid, which resembles a saddle.
This exercise involves an elliptic paraboloid given by the equation \(z = x^2 + 3y^2\). Here, the coefficients of \(x^2\) and \(y^2\) determine the stretching of the paraboloid along the \(x\)-axis and \(y\)-axis, respectively. The paraboloid intersects a plane to form a curve \(C\), and it is on this curve that the tangent line we need to find will be located.

Parametric Equations

Parametric equations are used to express geometric locations as functions of one or more variables, called parameters. Instead of representing a curve in Cartesian coordinates \((x, y, z)\), parametric equations use variables like \(t\) to describe the position coordinates in terms of one or more parametric equations:

• \(x(t)\) to describe the \(x\)-coordinate,
• \(y(t)\) for the \(y\)-coordinate,
• \(z(t)\) for the \(z\)-coordinate.

This exercise results in parametric equations for the tangent line, such as \(x(t) = 2 + ct, y(t) = 1 - ct, z(t) = 7 + ct\). These expressions effectively describe the line in three-dimensional space as \(t\), the parameter, varies.

Gradient Vector

The gradient vector is a critical component in multivariable calculus and vector analysis. It represents the direction and rate of fastest increase of a function at a given point. For a function \(f(x, y, z)\), the gradient (denoted as \(abla f\)) is composed of the partial derivatives of \(f\) with respect to each variable:

• \(\frac{\partial f}{\partial x}\)
• \(\frac{\partial f}{\partial y}\)
• \(\frac{\partial f}{\partial z}\)

In this exercise, the gradient of the paraboloid \(z = x^2 + 3y^2\) at the point \((2,1,7)\) is \(\langle 4, 6, -1\rangle\). This vector is normal to the tangent plane of the paraboloid at the given point, meaning it is perpendicular to any directional vector lying in that plane.

Direction Vector

A direction vector provides the direction in which a line or curve extends. In geometry and vector calculus, the direction vector is essential for constructing lines in space. It can be denoted as \(\langle a, b, c \rangle\) and indicates how much movement occurs in the \(x\), \(y\), and \(z\) directions respectively.
In this exercise, we identify the direction vector for the tangent line by ensuring it is perpendicular to both normal vectors of the intersecting planes, leading to equations like \(a + b - c = 0\) and \(4a + 6b - c = 0\). Solving these equations gives us a direction vector \(\langle c, -c, c \rangle\). This property ensures the tangent line lies along the curve formed by the intersection of the plane and the paraboloid.