Parametric Equations of Tangent Line

1. Mar 7, 2016

njo

1. The problem statement, all variables and given/known data
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)

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

3. 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: Mar 7, 2016
2. Mar 7, 2016

Staff: Mentor

Why does z appear on both sides? Is the equation actually w = 2x^2 + 5y^2 + z? Or is the last z a typo?

3. Mar 7, 2016

njo

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

That's the equation. Typo.

4. Mar 7, 2016

Staff: Mentor

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.

5. Mar 7, 2016

njo

dz/dy = 10y

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

Is this correct?

6. Mar 7, 2016

Staff: Mentor

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

7. Mar 7, 2016

njo

Oops, I know better than that.

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

8. Mar 7, 2016

Staff: Mentor

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.

9. Mar 8, 2016

geoffrey159

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 $10y - z + 5 = 0$.

10. Mar 8, 2016

njo

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?

11. Mar 8, 2016

Staff: Mentor

That's what I get as well.

12. Mar 9, 2016

njo

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