# Parametric Equations of Tangent Line

• njo
In summary: Therefore, in summary, for the given equation z = 2x^2 + 5y^2 +2, the line tangent to the curve C at the point (2, 1, 15) can be represented by the parametric equations y = t + 1 and z = 10t + 15.
njo

## Homework Statement

z = 2x^2 + 5y^2 +2
C is cut by the plane x = 2
Find parametric eqns of the line tangent to C @ P(2, 1, 15)

z = 5y^2 + 10
dz/dx = 10y
dz/dx (1) = 10

## The Attempt at a Solution

z = 10y + 15
y = t + 1

if the slope is 10/1 then delta z = 10 and delta y = 1. Therefore those values are multiples in front of t for the z and y parametric eqns respectively. Is this correct?

Last edited:
njo said:

## Homework Statement

z = 2x^2 + 5y^2 +z
Why does z appear on both sides? Is the equation actually w = 2x^2 + 5y^2 + z? Or is the last z a typo?
njo said:
C is cut by the plane x = 2
Find parametric eqns of the line tangent to C @ P(2, 1, 15)

z = 5y^2 + 10
dz/dx = 10y
dz/dx (1) = 10

## The Attempt at a Solution

z = 10y + 15
y = t + 1

if the slope is 10/1 then delta z = 10 and delta y = 1. Therefore those values are multiples in front of t for the z and y parametric eqns respectively. Is this correct?

z = 2x^2 + 5y^2 +2

That's the equation. Typo.

njo said:
z = 5y^2 + 10
dz/dx = 10y
That would be dz/dy, not dz/dx.

The tangent line is a line in ##\mathbb{R}^3##. Parametric equations would be of the form x = f(t), y = g(t), z = h(t). Since the tangent line is in the plane z = 2, x will be just 2.

dz/dy = 10y

z = 10y + 15
y = t + 1
x = 2

Is this correct?

njo said:
dz/dy = 10y

z = 10y + 15
y = t + 1
x = 2

Is this correct?
The equations for y and z should be in terms of a parameter. Typically, t is used.

Oops, I know better than that.

z = 10t + 15
y = t + 1
x = 2

njo said:
Oops, I know better than that.

z = 10t + 15
y = t + 1
x = 2
That's not what I get. The trace of the surface ##z = 2x^2 + 5y^2 + 2## in the plane x = 2 is ##z = 5y^2 + 10##, a parabola. The slope of the tangent line to this parabola at the point (2, 1, 15) is 10, which you have, but I get a different equation for the tangent line. Use the point-slope form of the equation of the line, with m = 10, and the point (1, 15) -- (y, z) coordinates.

You want to find the tangent line as the intersection of plane ## P_1 |\ x = 2 ## and plane ## P_2 |\ a x + by + cz + d = 0 ##,
where ## (a,b,c) = \text{grad }F(2,1,15) ##, ##F(x,y,z) = 2x^2 + 5y^2 -z + 2 ##, and ##d## determined by the fact that ##(2,1,15)## belongs to ##P_2##. In cartesian coordinates, I get that the tangent line is ## 10y - z + 5 = 0 ##.

So I find the gradient of w = 2x^2 + 5y^2 - z + 2 which would be <4x, 10y, -1>. I substitute P into the gradient? and that's how I get the coefficient in front of the parameter?

geoffrey159 said:
I get that the tangent line is ## 10y - z + 5 = 0 ##.
That's what I get as well.

The tangent line is z = 10y+5 therefore the parametric equations of this line are:
y = t +1
z = 10t + 15

## 1. What are parametric equations of tangent lines?

Parametric equations of tangent lines are equations that describe the relationship between a point on a curve and the slope of the tangent line at that point. These equations are typically expressed in terms of a parameter, such as time or distance.

## 2. How are parametric equations of tangent lines different from regular tangent line equations?

Parametric equations of tangent lines are different from regular tangent line equations because they take into account the changing slope of a curve at different points. Regular tangent line equations only give the slope at a specific point on a curve.

## 3. How are parametric equations of tangent lines used in real-life situations?

Parametric equations of tangent lines are used in real-life situations to model the rate of change of a variable over time or distance. For example, they can be used to study the trajectory of a projectile or the growth of a population.

## 4. What is the process for finding parametric equations of tangent lines?

The process for finding parametric equations of tangent lines involves finding the derivative of the given curve, plugging in the coordinates of the point of interest, and solving for the slope. The resulting slope is then used to create the tangent line equation.

## 5. Can parametric equations of tangent lines be used for any type of curve?

Yes, parametric equations of tangent lines can be used for any type of curve, including circles, ellipses, and more complex curves. As long as the curve is differentiable, parametric equations of tangent lines can be found and used to model the curve's behavior.

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