Lesson 2.16 — Mixed Differentiation, Tangent and Normal Lines

Thinking task: four digits

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2.16 · Mixed Differentiation · Tangent and Normal Lines

Use each digit 1, 2, 3, 4 exactly once. Make two two-digit numbers.

What is the smallest positive difference? Justify.

$31-24=7$

Tens contribute at least 10.

Ones can reduce that gap by at most 3.

$10-3=7$

Do-Now

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The functions are differentiable wherever needed.

$h(x)=f(x)g(x)+f(g(x))$

Input	$f$ value	$f$ slope	$g$ value	$g$ slope
2	1	$-3$	4	2
4	5	$\frac12$	—	—

Find $h(2)$ , $h'(2)$ , and the tangent line at input 2.

Choose each rule from the structure

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$h(2)=1\cdot4+5=9$

$h'(2)=f'(2)g(2)+f(2)g'(2)+f'(g(2))g'(2)$

$h'(2)=(-3)(4)+(1)(2)+\frac12(2)=-9$

$y-9=-9(x-2)$

The normal uses the same point

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A normal line passes through the point and is perpendicular to the tangent line.

$P=(a,f(a)),\quad m=f'(a)\ne0$

$\text{Tangent: }y-f(a)=m(x-a)$

What slope must the normal have?

$\text{Normal: }y-f(a)=-\frac1m(x-a)$

$m=-9\quad\Rightarrow\quad m_N=\frac19$

Three kinds of behavior at the origin

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$y=x^2$

Parabola y equals x squared at the origin.

$x=y^3$

Smooth curve x equals y cubed at the origin.

$y=|x|$

Absolute value curve with a corner at the origin.

At the origin, identify the tangent and normal, or explain why there is no unique line.

Undefined does not always mean vertical

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$y=x^2$

Parabola with a horizontal tangent and vertical normal.

$\text{T: }y=0\quad\text{N: }x=0$

$x=y^3$

Curve x equals y cubed with a vertical tangent and horizontal normal.

$\text{T: }x=0\quad\text{N: }y=0$

$y=|x|$

Absolute value curve; no unique tangent or normal at the corner.

Unequal one-sided slopes.

No single tangent or normal.

Solid teal: tangent · Dashed blue: normal

One function, two derivatives, two lines

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$q(x)=x^2\ln x,\quad x>0$

Find the first and second derivatives.

At input 1, find the tangent and normal lines.

Keep the value and the slopes distinct

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$q'(x)=2x\ln x+x^2\frac1x=2x\ln x+x$

$q''(x)=2\ln x+2x\frac1x+1=2\ln x+3$

$q(1)=0,\quad q'(1)=1,\quad q''(1)=3$

$\text{T: }y=x-1\qquad\text{N: }y=1-x$

C1 · Read the geometry backward

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At every point with positive input, the tangent meets the $y$ -axis exactly one unit below the point.

$P=(a,f(a)),\quad a>0$

For what value of $a$ are the $y$ -intercepts of the tangent and normal 10 units apart?

Translate the tangent condition

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$y-f(a)=f'(a)(x-a)$

$x=0\quad\Rightarrow\quad b_T=f(a)-af'(a)$

$f(a)-af'(a)=f(a)-1$

$af'(a)=1\quad\Rightarrow\quad f'(a)=\frac1a$

Now compare the two intercepts

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$m_N=-\frac1{f'(a)}=-a$

$y-f(a)=-a(x-a)\quad\Rightarrow\quad b_N=f(a)+a^2$

$b_N-b_T=(f(a)+a^2)-(f(a)-1)=a^2+1$

$a^2+1=10,\quad a>0\quad\Rightarrow\quad a=3$

C2 · Does centered mean normal?

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$\frac{x^2}{16}+\frac{y^2}{9}=1$

“Every normal passes through the origin because the curve is centered there.”

Decide whether the claim is true.

Find every point where it is true.

Ellipse with semiaxes 4 and 3 centered at the origin, without solution lines.

First consider a point off both axes

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$\frac{x}{8}+\frac{2y}{9}y'=0\quad\Rightarrow\quad y'=-\frac{9x}{16y}$

$xy\ne0:\quad m_N=\frac{16y}{9x},\quad m_{OP}=\frac yx$

$\frac{16y}{9x}=\frac yx\quad\Rightarrow\quad16y=9y\quad\Rightarrow\quad y=0$

Contradiction: we assumed the point was off both axes.

Axis points complete the argument

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Ellipse with tangent and normal lines at the four vertices.

$\frac{dx}{dy}=-\frac{16y}{9x}=0\text{ at }(\pm4,0)$

$\text{T: }x=\pm4,\quad\text{N: }y=0$

$y'=0\text{ at }(0,\pm3)$

$\text{T: }y=\pm3,\quad\text{N: }x=0$

$(\pm4,0),\quad(0,\pm3)$

What changes for a circle?

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$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\quad a,b>0$

When does every normal pass through the center?

$xy\ne0:\quad m_N=\frac{a^2}{b^2}\frac yx$

Equal semiaxes: every point.

Unequal semiaxes: four vertices.

Exit ticket · What does the number prove?

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$f(x)=\begin{cases}x^2+1,&x\le1\\3x-1,&x>1\end{cases}$

A symmetric numerical routine reports approximately 2.5:

$D_h=\frac{f(1+h)-f(1-h)}{2h}$

Does the derivative at input 1 exist?

Is the tangent vertical? Justify.

Decide first. Differentiate. Check the line.

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The point locates the line.

The derivative gives the tangent slope.

Perpendicularity determines the normal.

Homework 2.16 — Mixed Differentiation, Tangent and Normal Lines

Questions 1–6, including 6*. No calculator. Show all work for full credit.

Optional explorations: use the Live graphs menu after the lesson.

A moving point, tangent, and normal

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Move the point. Predict the tangent before comparing the normal.

Static print state: $t=0.40$ ; tangent shown; normal hidden.

Test the ellipse claim

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When do the normal and the line through the center coincide?

Static print state: $\theta=45^\circ$ ; tangent shown; normal and radial segment hidden.

Your original Desmos graph

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Move the slider in your saved calculator. Open Desmos in a separate tab ↗

The original Desmos calculator loads here.

Internet required. The two programmed explorations work offline.

Static view of the folium x cubed plus y cubed equals 2xy with the point and tangent from the saved Desmos state at t equals negative 7.79.

Static print state: $t=-7.79$ , with the saved point and tangent. Open the original interactive graph.