Tangent Line to a Curve

Key Questions

How does tangent slope relate to the slope of a line?

The "tangent slope" is the slope of the tangent line. It is also called "the slope of the tangent" and "the slope of the curve at a point".

Jim H · 2 · Apr 13 2015
How do I find the equation for a tangent line without derivatives?

Answer:

You could use infinitesimals...

Explanation:

The slope of the tangent line is the instantaneous slope of the curve. So if we increase the value of the argument of a function by an infinitesimal amount, then the resulting change in the value of the function, divided by the infinitesimal will give the slope (modulo taking the standard part by discarding any remaining infinitesimals).

For example, suppose we want to find the tangent to $f (x)$ $f(x)$ at $x = 2$ $x=2$ , where:

$f (x) = x^{3} - 3 x^{2} + x + 5$ $f(x) = x^3-3x^2+x+5$

Let $ε > 0$ $epsilon > 0$ be an infinitesimal value. Then:

$\frac{f (2 + ε) - f (2)}{ε}$ $(f(2+epsilon) - f(2))/epsilon$

$= \frac{({(2 + ε)}^{3} - 3 {(2 + ε)}^{2} + (2 + ε) + 5) - ({(2)}^{3} - 3 {(2)}^{2} + (2) + 5)}{ε}$ $=(((2+epsilon)^3-3(2+epsilon)^2+(2+epsilon)+5)-((2)^3-3(2)^2+(2)+5))/epsilon$

$= \frac{((8 + 12 ε + 6 ε^{2} + ε^{3}) - 3 (4 + 4 ε + ε^{2}) + (2 + ε) + 5) - (8 - 12 + 2 + 5)}{ε}$ $=(((8+12epsilon+6epsilon^2+epsilon^3)-3(4+4epsilon+epsilon^2)+(2+epsilon)+5)-(8-12+2+5))/epsilon$

$= \frac{(12 ε + 6 ε^{2} + ε^{3}) - (12 ε + 3 ε^{2}) + ε}{ε}$ $=((12epsilon+6epsilon^2+epsilon^3)-(12epsilon+3epsilon^2)+epsilon)/epsilon$

$= \frac{ε + 3 ε^{2} + ε^{3}}{ε}$ $=(epsilon+3epsilon^2+epsilon^3)/epsilon$

$= 1 + 3 ε + ε^{2}$ $=1+3epsilon+epsilon^2$

of which the standard (i.e. finite) part is $1$ $1$ (discarding the $3 ε + ε^{2}$ $3epsilon+epsilon^2$ ).

So the slope of the tangent is $1$ $1$ and the tangent point is:

$(2, f (2)) = (2, 3)$ $(2, f(2)) = (2, 3)$

So the equation of the tangent may be written:

$(y - 3) = 1 (x - 2)$ $(y-3) = 1(x-2)$

or more simply:

$y = x + 1$ $y = x+1$

graph{ (y-(x^3-3x^2+x+5))(y-x-1) = 0 [-3.355, 6.645, 1.38, 6.38]}

George C. · 1 · Aug 4 2018
What is the difference between a Tangent line and a secant line on a curve?

The **tangent line ** to a curve at a given point is a straight line that just "touches" the curve at that point.

So if the function is f(x) and if the tangent "touches" its curve at x=c, then the tangent will pass through the point (c,f(c)). The slope of this tangent line is f'(c) ( the derivative of the function f(x) at x=c).

A secant line is one which intersects a curve at two points.

![)

Click this link for a detailed explanation on how calculus uses the properties of these two lines to define the derivative of a function at a point.

Avanika · 4 · Aug 17 2014

Questions

Derivatives

View all chapters

Science

Math

Humanities

... and beyond

Tangent Line to a Curve

Key Questions

Answer:

Explanation:

Questions

Derivatives