# Tangent line, to a surface, through a point, parallel to a plane

• crc1559
The parameter t can be replaced by any other parameter, such as x or z, to get the other two equations. For x= 1, z= 4x- 10. For z= 4z- 6, y= (z+ 6)/4.
crc1559
Hey y'all, this is my first post. I am currently stuck on a multivariable question. Please let me know if you can help.

## Homework Statement

The point, P = (1, 2, 2) lies on the surface z = x^2 + y^2 -3x. Find parametric equations for the tangent line to the surface through the point P parallel to the plane x = 1.

## Homework Equations

Gradient vector ∇F(x,y) = < dF/dx, dF/dy>
Normal vector n = < dF/dx, dF/dy, -1>

General form of tangent vector:
dF/dx(x-x0) + dF/dy(y-y0) + dF/dz(z-z0)

## The Attempt at a Solution

∇F(x,y) = < 2x -3, 2y >
n = < 2x-3, 2y, -1>
n(1, 2, 2) = < -1, 4, -1>

-1(x-1) + 4(y-2) -1(z-2) = 0
-x + 1 +4y - 8 -z + 2 = 0
-x + 4y -z = 5

This is where I am stuck.

In order to be parallel to the plane x=1, should the parametric equations be
x = 1, y = 2 + 4t, z = 2 - t
or would it still be the tangent equation
x = 1 - t, y = 2 + 4t, z = 2-t?

crc1559 said:
Hey y'all, this is my first post. I am currently stuck on a multivariable question. Please let me know if you can help.

## Homework Statement

The point, P = (1, 2, 2) lies on the surface z = x^2 + y^2 -3x. Find parametric equations for the tangent line to the surface through the point P parallel to the plane x = 1.

## Homework Equations

Gradient vector ∇F(x,y) = < dF/dx, dF/dy>
Normal vector n = < dF/dx, dF/dy, -1>

General form of tangent vector:
dF/dx(x-x0) + dF/dy(y-y0) + dF/dz(z-z0)
This is not a tangent vector (or any vector at all). You cannot simply get an equation for "the" tangent vector to a surface at a point because there exist an infinite number of tangent vectors at a given point. I think what you mean is that the tangent plane satisfies the equation $(\partial F/\partial)(x- x_0)+ (\partial F/\partial y)(y- y_0)+ (\partial F/\partial z)(z- z_0)= 0$. For this exercise that is -(x- 1)+ 4(y- 2)- (z- 2)= 0 or x- 4y- z= 5. You want a line in that plane and in the plane x=1. That is, the line of intersection of the two planes.

## The Attempt at a Solution

∇F(x,y) = < 2x -3, 2y >
n = < 2x-3, 2y, -1>
n(1, 2, 2) = < -1, 4, -1>

-1(x-1) + 4(y-2) -1(z-2) = 0
-x + 1 +4y - 8 -z + 2 = 0
-x + 4y -z = 5

This is where I am stuck.

In order to be parallel to the plane x=1, should the parametric equations be
x = 1, y = 2 + 4t, z = 2 - t
or would it still be the tangent equation
x = 1 - t, y = 2 + 4t, z = 2-t?
Where did you get these equations from? The first set, x= 1, y= 2+4t, z= 2- t does not satisfy -x+ 4y- z=5: -1+4(2+ 4t)- (2- t)= -1+ 8+ 16t- 2+ t= 5+ 17t, not 5 so does not lie in the tangent plane. The second does not have x always equal to 1 so does not lie in the plane x= 1 nor is it parallel to it.

Since x= 1 at the given point, the plane x=1 is through the given point and "parallel to the plane x= 1" means "lies in the plane x= 1" so the x coordinate is the constant, 1. Since you have correctly determined that the tangent plane is given by -x+ 4y- z= 5, the intersection of the two planes gives -1+ 4y- z= 5 or z= 4y- 6. Taking y itself as parameter, the line of intersection is x= 1, y= t, z= 4t- 6.

## 1. What is a tangent line?

A tangent line is a line that touches a curve or surface at only one point. It is perpendicular to the curve or surface at that point, and it represents the slope of the curve or surface at that point.

## 2. How is a tangent line different from a secant line?

A secant line is a line that intersects a curve or surface at two different points, while a tangent line only touches the curve or surface at one point. The slope of a secant line represents the average rate of change over an interval, while the slope of a tangent line represents the instantaneous rate of change at a specific point.

## 3. What does it mean for a tangent line to be parallel to a plane?

When a tangent line is parallel to a plane, it means that the tangent line and the plane do not intersect. This also means that the slope of the tangent line is equal to the slope of any line on the plane that is parallel to the tangent line.

## 4. How is a tangent line to a surface calculated?

A tangent line to a surface is calculated by finding the partial derivatives of the surface at the given point. These partial derivatives represent the slopes in the x and y directions, and the tangent line can then be determined using these slopes at the given point.

## 5. Why is a tangent line important in calculus?

In calculus, the tangent line is important because it helps us understand the behavior of functions and surfaces at a specific point. It allows us to find the instantaneous rate of change and also helps us approximate the behavior of a function or surface near that point. Tangent lines also play a crucial role in related rates problems and optimization problems in calculus.

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