Menu
Home
Action
My entries
Defined browse
Select

Then Select

Then Select

Search

 

tangent line


Definition/Summary
The tangent to a curve in a plane at a particular point has the same Gradient as the curve has at that point.

More generally, the (n-1)-dimensional tangent hyperplane to an (n-1)-dimensional surface in n-dimensional space at a particular point has the same Gradient as the surface has at that point.

So if [itex]A\,=\,(a_1,a_2,\cdots a_n)[/itex] is a point on a surface defined by the equation [itex]F(x_1,x_2,\cdots x_n) = 0[/itex], then the tangent hyperplane to the curve through [itex]A[/itex] is [itex]\frac{\partial F}{\partial x_1}\arrowvert_A(x_1 - a_1)\,+\,\frac{\partial F}{\partial x_2}\arrowvert_A(x_2 - a_2)\,+\,\cdots\,\frac{\partial F}{\partial x_2}\arrowvert_A(x_n - a_n)\,=\,0[/itex]

If a curve in n dimensions is defined using a parameter t as [itex]A(t)\,=\,(a_1(t),a_2(t),\cdots a_n(t))[/itex] , then its tangent is:
[itex](x_1 - a_1) / \frac{da_1}{dt}\,=\,(x_2 - a_2) / \frac{da_2}{dt}\,=\,\cdots\,=\,(x_n - a_n) / \frac{da_n}{dt}[/itex]

Equations
For example, if the point [itex]A\,=\,(x_0,y_0)[/itex] lies on the circle:

[tex]F(x,y)\,=\,(x-p)^2+(y-q)^2\,-\,r^2\,=\,0[/tex] (1)

then [tex]\frac{\partial F}{\partial x}\arrowvert_A\,=\,2(x-p)\arrowvert_A\,=\,2(x_0-p)[/tex]
and [tex]\frac{\partial F}{\partial y}\arrowvert_A\,=\,2(y-q)\arrowvert_A\,=\,2(y_0-q)[/tex]

and so the equation of the tangent at [itex]A[/itex] is:

[tex](x_0-p)(x-x_0)\,+\,(y_0-q)(y-y_0)=0[/tex] (2)

Alternatively, the same circle can be defined by [itex]A(\theta)\,=\,(p+r\cos\theta, q+r\sin\theta)[/itex]
and so the equation of the tangent at [itex]A(\theta)[/itex] is:

[tex]\frac{x-p-r\cos\theta}{-r\sin\theta}\,=\,\frac{y-q-r\sin\theta}{r\cos\theta}[/tex]

Scientists

Recent forum threads on tangent line
 
Breakdown
Mathematics
> Geometry
>> Coordinate Geometry

See Also
Gradient

Images
Click to enlarge


Extended explanation
We will transform the equation (2) into more convenient type for better way of memorizing and using the formula. Because of [itex]M(x_1,y_1) /in K[/itex]:

[tex](x_1-p)+(y_1-q)^2=r^2[/tex] (3)

If we sum the equations (2) and (3), we get:

[tex](x_1-p)(x-x_1)+(y_1-q)(y-y_1)+(x_1-p)+(y_1-q)^2=r^2[/tex]

[tex](x_1-p)[x-x_1+x_1-p]+(y_1-q)[y-y_1+y_1-q]=r^2[/tex]

[tex](x_1-p)(x-p)+(y_1-q)(y-q)=r^2[/tex] (4)

The equation (4) is equation of tangent of the circle in the point [itex]M(x_1,y_1) \in K[/itex].

If the K have center (0,0), i.e [itex]K: x^2+y^2=r^2[/itex], then p=q=0, so the equation of the tangent is:

[tex]x_1x+y_1y=r^2[/tex]

Commentary

tiny-tim @ 01:59 PM Jan16-09
changed title from 'tangent to a curve' to 'tangent line' because a forum search shows that that will give 20 times as much autolinking