# Thread: Equation of tangent line to curve

1. ## Equation of tangent line to curve

The curve r(t) passing through the point (1,1,1) is defined implicitly by the equations:

x^2y + y^2x + xyz + z^2 = 4
x + y + z = 3

Find the equation of the tangent line to the curve at the given point.

Attempt at a solution

GRAD f(x,y,z) = (2xy + y^2)i + (x^2 + 2xy)j + (xy + 2z)k

At (1,1,1):

GRAD f(1,1,1) = 3i + 3j + 3k

Tangent line has normal of (3,3,3).

Therefore:

3(x - 1) + 3(y - 1) + 3(z - 1) = 0
3x + 3y + 3z = 9
x + y + z = 3

I'm not sure what I'm doing wrong. Am I supposed to use both equations given in the question?

2. Accidental double post.

3. Originally Posted by My Little Pony
The curve r(t) passing through the point (1,1,1) is defined implicitly by the equations:

x^2y + y^2x + xyz + z^2 = 4
x + y + z = 3

Find the equation of the tangent line to the curve at the given point.

Attempt at a solution

GRAD f(x,y,z) = (2xy + y^2)i + (x^2 + 2xy)j + (xy + 2z)k

At (1,1,1):

GRAD f(1,1,1) = 3i + 3j + 3k

Tangent line has normal of (3,3,3).

Therefore:

3(x - 1) + 3(y - 1) + 3(z - 1) = 0
3x + 3y + 3z = 9
x + y + z = 3

I'm not sure what I'm doing wrong. Am I supposed to use both equations given in the question?
Since the curve is formed by both surfaces, don't you think that would be a good idea? A single curve does not define a surface, much less its tangent line.

But you started correctly: (3,3,3) is normal to the first surface and so is normal to any line in the surface. But your answer x+ y+ z= 3 is the equation of the plane tangent to the surface there.

The difficulty with using just one surface is that there exist an infinite number of vectors perpendicular to a give vector. There exist and infinite number of tangent lines to that surface.

Instead, find the normal to the second surface (which is actually a plane) at (1, 1, 1). The vector you want must be perpendicular to both of those- in other words it is their cross product.

Now that I look at it more closely, I see a problem with that: the two given surfaces are tangent at (1, 1, 1). The don't intersect in a curve to begin with! There is NO curve "defined implicitly" by those equations.

4. Originally Posted by My Little Pony
The curve r(t) passing through the point (1,1,1) is defined implicitly by the equations:

x^2y + y^2x + xyz + z^2 = 4
x + y + z = 3

Find the equation of the tangent line to the curve at the given point.

Attempt at a solution

GRAD f(x,y,z) = (2xy + y^2)i + (x^2 + 2xy)j + (xy + 2z)k

At (1,1,1):

GRAD f(1,1,1) = 3i + 3j + 3k

Tangent line has normal of (3,3,3).

Therefore:

3(x - 1) + 3(y - 1) + 3(z - 1) = 0
3x + 3y + 3z = 9
x + y + z = 3

I'm not sure what I'm doing wrong. Am I supposed to use both equations given in the question?
Originally Posted by HallsofIvy
. . .
Now that I look at it more closely, I see a problem with that: the two given surfaces are tangent at (1, 1, 1). The don't intersect in a curve to begin with! There is NO curve "defined implicitly" by those equations.
I think you do. If you eliminate $z$ using the plane, then you have

$
x^2y + y^2x + xy(3-x-y) + (3-x-y)^2 = 4
$

or

$
3xy + (3-x-y)^2 = 4\;\;\;(1)
$

which is the curve projected into the $xy$ plane.

@ little pony. As for the tangent line. It you let $x = t$, (so $\frac{dx}{dt} = 1$) then from (1) you can find $\frac{dy}{dt}.$ For the last $\frac{dz}{dt} = - \frac{dx}{dt} - \frac{dy}{dt}$ so now you have the three conponents of the tangent vector. You use the point $(1,1,1)$ to get the tangent line.

5. Originally Posted by HallsofIvy
Now that I look at it more closely, I see a problem with that: the two given surfaces are tangent at (1, 1, 1). The don't intersect in a curve to begin with! There is NO curve "defined implicitly" by those equations.
Ah sorry, I believe I computed the gradient incorrectly.

The normal of the first equation should actually be (4,4,3).

Thanks a bunch for your help!