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.

Short answer

The tangent line to the curve at the point (2,1,7) is given by the equation \(\mathbf{r}(t) = \langle 2+t, 1-t, 7+2t \rangle\), where the direction vector is determined by orthogonal normal vectors to both the given plane and tangent plane to the paraboloid.

Step-by-step solution

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

To find a normal vector to the plane, we need to find the gradient of the plane equation. The equation of the plane is given as \(z=x+y+4\). The gradient of this equation is the coefficients of the \(x\) and \(y\) terms, which are 1 and 1. Therefore, a normal vector to the plane is \(\mathbf{n}_{1}=\langle1,1,0\rangle\).

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

The equation of the paraboloid is given as \(z=x^{2}+3y^{2}\). To find a normal vector to the plane tangent to the paraboloid, we need the gradient of the paraboloid equation. The partial derivatives of the paraboloid with respect to \(x\) and \(y\) are: \(\frac{\partial z}{\partial x}=2x\) \(\frac{\partial z}{\partial y}=6y\) Now, we find the gradient at the point (2,1,7) by evaluating the partial derivatives at the given point: \(\frac{\partial z}{\partial x}(2,1,7)=4\) \(\frac{\partial z}{\partial y}(2,1,7)=6\) So, the normal vector to the tangent plane of the paraboloid at (2,1,7) is \(\mathbf{n}_{2}=\langle4,6,1\rangle\).

Show that the tangent line to the curve is orthogonal to both normal vectors

When two vectors are orthogonal, their dot product is 0. In our problem, it means that the tangent line must be orthogonal to both normal vectors. Let's denote the direction vector of the tangent line as \(\mathbf{v}=\langle a,b,c \rangle\). For \(\mathbf{v}\) to be orthogonal to both normal vectors, their dot product must be zero: \(\mathbf{v}\cdot \mathbf{n}_{1}=a + b + 0 = 0\) \(\mathbf{v}\cdot \mathbf{n}_{2}=4a + 6b + c = 0\)

Use this fact to determine a direction vector for the tangent line

We have the following system of linear equations from Step 3: \(a + b = 0\) \(4a + 6b +c = 0\) Since we only need a direction vector, we can choose an arbitrary value for one variable and solve the system of equations in terms of the chosen variable. Let's set \(a=1\), so \(b=-1\). Substituting these values into the second equation: \(4(1) + 6(-1) + c = 0\) \(c= 2\) Thus, the direction vector of the tangent line is \(\mathbf{v}=\langle 1,-1,2\rangle\). Finally, to find the equation of the tangent line, we can use the point-slope form: \(\mathbf{r}(t) = \mathbf{P}_{0} + t\mathbf{v}\) Substitute the given point (2,1,7) and the direction vector \(\mathbf{v}=\langle 1,-1,2\rangle\): \(\mathbf{r}(t) = \langle 2,1,7 \rangle + t \langle 1,-1,2 \rangle\) Therefore, the equation of the tangent line to the curve at the point (2,1,7) is: \(\mathbf{r}(t) = \langle 2+t, 1-t, 7+2t \rangle\).

Key concepts

Normal Vector

A normal vector is an essential concept in the world of vectors and geometry. It is a vector that is perpendicular to a surface or a line. In the context of a two-dimensional plane, suppose we have a line. The normal vector to this line would be a vector that points directly away from the line at a 90-degree angle.

In three-dimensional space, this concept gets even more interesting. If you have a surface, like a plane or the side of a paraboloid, a normal vector to that surface at a specific point is a vector that sticks out perpendicularly from that spot. For example, the solution shows that to find the normal vector \(\mathbf{n}_{1}\) to the plane at the point (2,1,7), we look at the coefficients of \(x\) and \(y\) in the plane's equation and form a vector with them, resulting in \(\mathbf{n}_{1}=\langle1,1,0\rangle\).

Every surface has an infinite number of normal vectors at any given point, all extending out in the same perpendicular direction but with different lengths. The normal vector is a fundamental building block when solving for tangents, as seen in our textbook exercise.

Gradient of a Function

The gradient of a function is like a compass pointing uphill on the terrain of the function's graph. In more technical terms, the gradient is a vector containing all the partial derivatives of a function, and it gives the direction of the steepest ascent.

Here's a way to picture it: suppose you're hiking on a hill that the function describes. The gradient tells you the steepest way up at any given point. So, in our problem, when we find the gradient of the function for the paraboloid \(z=x^{2}+3y^{2}\), we calculate the partial derivatives with respect to \(x\) and \(y\) and put them into a vector. This gives us the gradient at any point, which also serves as the normal vector to the tangent plane at that point on the surface. Understanding this idea is critical when you're trying to find tangent lines or planes in multivariable calculus.

Partial Derivatives

When dealing with functions of more than one variable, we often want to see how the function changes as we adjust one variable at a time - this is where partial derivatives come in. They are like regular derivatives, but they focus on change in only one dimension, holding the other variables constant.

In the exercise, we calculate the partial derivatives \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\) for the paraboloid's equation. These derivatives tell us how fast the \(z\)-value of the paraboloid changes as we move in either the \(x\)- or \(y\)-directions, while keeping the other variable fixed. They help form the gradient and hence the normal vector of the function at a given point. Getting to grips with partial derivatives is essential for anyone diving into topics like tangent planes, optimization problems, or any application involving multivariable functions.

Dot Product

The dot product, also known as the scalar product, is a way to multiply two vectors that gives us a scalar, or a single number. This number tells us something about the relationship between the vectors, particularly, how they line up with each other. If the dot product is zero, it means the vectors are orthogonal, or perpendicular.

For instance, in our exercise, to confirm that the tangent line is perpendicular to both normal vectors, we use the dot product. We set up equations \(\mathbf{v}\cdot \mathbf{n}_{1}=0\) and \(\mathbf{v}\cdot \mathbf{n}_{2}=0\), where \(\mathbf{v}\) is the direction vector of the tangent line. A zero dot product tells us we've found our perfect tangent - one that doesn't lean even slightly in the direction of these normal vectors.

Direction Vector

The concept of a direction vector is like having coordinates for an arrow that points 'this way.' It tells us the direction in which a line or vector is heading. Specifically, a direction vector has the same direction as a line but doesn't necessarily share the same length or starting point.

In the context of our textbook problem, we strive to find a direction vector that is orthogonal to two given normal vectors, which guarantees it is tangent to the curve at the specified point. As the solution reveals, we solve a system of linear equations derived from the dot product condition. This led to the discovery of the direction vector \(\mathbf{v}=\langle 1,-1,2\rangle\) for the tangent line. Understanding direction vectors is vital as it allows students to navigate within the realm of lines and curves in multidimensional space.