Line Tangent to a Curve

1. Overview. Suppose we have a line {\ell} in the plane {{\mathbb R}^{2}}. What data do we need to specify a line?

We need a slope {m} and a point {(x_{0}, y_{0})} on the line {\ell}. Then we can specify {\ell} completely by the equation
(1)\displaystyle  \frac{y-y_{0}}{x-x_{0}}=m.
We can rewrite this as
(2)\displaystyle  y = y_{0} + m(x-x_{0}).
Lets remember this.

Last time we discussed the derivative, and the derivative gives us the slope at a point {(x,f(x))}. But look: we have the slope from the derivative.

What more information do we need to compute a line? We need a point on the curve! So if {y=f(x)} is a curve, and {(x_{0},y_{0}=f(x_{0}))} is a point on the curve, then we can compute the line tangent to {y} at {x_{0}} as
(3)\displaystyle  t_{y_{0}}(x) = y_{0} + \left.\frac{\mathrm{d} f}{\mathrm{d} x}\right|_{x=x_{0}}\cdot(x-x_{0})
This uses the derivative {f'(x_{0})} as the slope, and the data {(x_{0}, y_{0})}.

(We should quickly prove to ourselves this passes through the desired point. It is a one minute check that t_{y_{0}}(x_{0}) = y_{0}+0. We also see it has the desired slope, and it is in fact a line.)

Lets now consider a few examples.

2. Basic Algorithm. So, what’s the basic algorithm when finding the tangent line of the curve y=f(x) at the point (x_{0}, y_{0})?

Step 1. Find the derivative f'(x).
Step 2. Using Step 1’s results, evaluate the derivative at x_{0}, i.e., evaluate f'(x_{0}).
Step 3. Using Step 2’s results, plug it into the formula t(x)=y_{0}+f'(x_{0})\cdot(x-x_{0}).

We will do each step and draw the doodles as we go.

Example 1. Lets find the tangent line passing through the point (1,2) for the curve y=f(x)=x^{2}+1. What to do?

Step 0: draw the curve. We will doodle the curve in blue, just so we understand which part of the graph is the curve.

Step 1: find the derivative f'(x). We see that the derivative for x^{2}+1 is, well, by using linearity
Then we invoke the power rule to find
\displaystyle\frac{\mathrm{d}(x^{2}}{\mathrm{d}x}+\frac{\mathrm{d}(1)}{\mathrm{d}x}=2x + 0
Thus we find
\displaystyle f'(x)=2x
This concludes step 1.

Step 2: Evaluate f'(x_{0}). OK, so we just found f'(x)=2x and we have x_{0}=1. Thus we find
\displaystyle f'(1)=2
This concludes step 2.

Step 3: Construct Tangent Line. Well, we have the formula for the tangent line be
\displaystyle t(x)=y_{0}+f'(x_{0})\cdot(x-x_{0})
We have everything we need. We plug in y_{0}=2 and x_{0}=1, and results from step 2, to get
\displaystyle t(x)=2+2\cdot(x-1)
But we may simplify things out:
\displaystyle t(x)=2+2x-2=2x
Thus the tangent to f(x)=x^{2}+1 at (1,2) is
\displaystyle t(x)=2x
We can doodle this:

Note the curve is doodled in blue, the tangent point is emphasized, and the tangent line is doodled in pink.

We will conclude this post with motivation for the next:

Problem. We want to find some curve {y=f(x)} such that at any point {(x_0,y_0)} on the curve its tangent line {t_{y_{0}}} satisfies the property {t_{y_{0}}(x_{0}-1)=0}. What is {y=f(x)}?

(The field called “Differential Equations” studies this general sort of problem “Find the curve whose derivative satisfies…”)


About Alex Nelson

I like math. I like programming. Most of all, I love puzzles.
This entry was posted in Calculus, Differential, Tangent Line. Bookmark the permalink.

5 Responses to Line Tangent to a Curve

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