Tangent Line Approximation

The Tangent Line Approximation – HMC Calculus Tutorial

Suppose we want to find the tangent to a curve. Just how can we go about finding one?

Here is one way:

  • Pick a point $Q$ by clicking on the curve on the applet, which is the line that appears is the secant line between $P$ and $Q$.

  • Now drag point $Q$ towards point $P$.

As $Q$ approaches $P$, the secant line approximates the tangent line better and better. The limiting position of the secant line as $Q$ approaches $P$ is the tangent to the curve at $P$.

If the curve is given by $y=f(x)$ and $P$ has the coordinates $(x_0,y_0)$, then the slope of the tangent line at $P$ is $f'(x_0)$, the derivative of f evaluated at $x_0$.

Let’s find the equation of the line tangent to the parabola at $(2,3)$.

  • Drag point $P$ to $(2,3)$.

  • Now pick another point $Q$ on the parabola and drag $Q$ towards $P$ to find the tangent to the curve at $P$.

The slope of the tangent is just $f'(x)$ evaluated at x. \begin{eqnarray*} f(x) &=& x^2-1 \\ f'(x) &=& 2x \\ f'(2) &=& 4. \end{eqnarray*} Now, the equation of the line can be written in point-slope form. The point-slope form of the equation of the line through the point $(x_0,y_0)$ with slope $m$ is given by \[y-y_0 = m(x-x_0).\]

So we can write like this: \begin{eqnarray*} y-y_0 &=& m(x-x_0)\\ y-y_0 &=& f'(x_0)(x-x_0)\\ y-3 &=& 4(x-2) \end{eqnarray*} since the line passes through the point $(2,3)$ and has slope $4$.

In slope-intercept form, the line with slope $m$ and $y$-intercept $b$ is given by \[y = mx+b.\], so the equation of the tangent line becomes $$ y=4x-5. $$

  • Drag $P$ along the parabola or enter the x-coordinate for point $P$.

  • Notice how the equation of the tangent line changes as you move point $P$.

What happens when $x=0$ for this function? What about as $|x|$ gets large?

Now that we can find the tangent to a curve at a point, of what use is this?

  • “Magnify” the parabola by zooming in on point $P$.

Do you notice that as you zoom in on $P$ the curve looks more and more linear and is approximated better and better by the tangent line?

Let’s get more specific:

Near $x_0$, we saw that $y=f(x)$ can be approximated by the tangent line $y-y_0=f'(x_0)(x-x_0)$. Writing this as $y=y_0+f'(x_0)(x-x_0)$ and noting that $y=f(x_0)$, we find that

$f(x)\approx f(x_0) + f'(x_0)(x-x_0).$

Suppose $f$ can be differentiated $n+1$ times at each point in some interval containing $x_0$. Then for $x$ in this interval about $x_0$, $$ f(x) \approx f(x_0) + f'(x_0)(x-x_0) + \frac{f”(x_0)}{2!}(x-x_0)^2 + \dots + \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n. $$. Notice that the right-hand side is just the 2-term Taylor Expansion of $f(x)$.

If we know that value of $f$ at $x_0$, this gives us a way to approximate the value of $f$ at $x$ near $x_0$. We do this by starting at $(x_0,f(x_0))$ and moving along the tangent line to approximate the value of the function at $x$.

Look at $f(x) = \arctan{x}$.

Let’s use the tangent approximation $f(x) \approx f(x_0)+f'(x_0)(x-x_0)$ to approximate $f(1.04)$:

  • Now $f'(x) = \left[\frac{1}{1+x^2}\right]$ so $f'(1)=\left[\frac{1}{1+1^2}\right]=\frac{1}{2}$.

  • Let $x_0 = 1$ and $x = 1.04$.

  • Then $f(1.04) \approx f(1) + f'(1)(1.04 – 1) \approx \frac{\pi}{4} + \frac{1}{2}(0.04) \approx 0.81$.

How well does this approximate $\arctan(1.04)$?

  • Display the tangent through $\left( 1, \frac{\pi}{4}\right)$.

  • Zoom in on the point to see geometrically how close together the curve and the tangent line are at $x = 1.04$.

Key Concepts

  • For the curve $y = f(x)$, the slope of the tangent line at a point $(x_0,y_0)$ on the curve is $f'(x_0)$. The equation of the tangent line is given by $$ y-y_0 = f'(x_0)(x-x_0). $$

  • For $x$ close to $x_0$, the value of $f(x)$ may be approximated by $$ f(x) \approx f(x_0)+f'(x_0)(x-x_0). $$

[I’m ready to take the quiz.] [I need to review more.]