Chapter 3 · Applications of Derivatives

3.1 Tangent and Normal Lines

The derivative $f'(a)$ is a slope. Put it through the point $(a,f(a))$ and you have the tangent line; turn it through a right angle and you have the normal line.

In Chapter 2 we learned to compute derivatives, and we saw that $f'(a)$ is the slope of the tangent line to the graph of $f$ at the point $(a,f(a))$. In this section we turn those slopes into lines: we write the equations of tangent and normal lines, find the points where a tangent is horizontal or has a prescribed direction, and do the same for curves given implicitly or parametrically. A line is the simplest curve there is, so replacing a curve by its tangent line near a point is one of the most useful ideas in applied mathematics.

🌍 Where tangent and normal lines appear
  • Estimation and linearisation. Near $x=a$ a curve is almost indistinguishable from its tangent line, so the tangent gives quick estimates such as $\sqrt{4.1}\approx2.025$. Engineers linearise the response of a sensor (a thermistor, a strain gauge) about its operating point in exactly this way.
  • Leaving a curved path. A stone released from a sling, or a car that skids off a bend, continues in the direction of the tangent to its path at the moment of release — not outward along the radius.
  • Headlights on a curved road. A car's headlights point along the tangent to the road, which is why they sweep across the verge on a bend.
  • Optics. Angles of incidence and reflection at a curved mirror or lens are measured from the normal line at the point where the ray strikes the surface.
  • Forces and design. The normal force of a road or rail on a vehicle acts along the normal to the surface; road and railway engineers shape bends so that tangent directions change smoothly.

The Equation of the Tangent Line

📘 Definition — Tangent line
Let $f$ be differentiable at $a$. The tangent line to the graph of $f$ at $a$ is the line through the point $(a,f(a))$ with slope $m=f'(a)$. In point-slope form its equation is $$y=f(a)+f'(a)(x-a).$$ Two ingredients are always needed: the point $(a,f(a))$ and the slope $f'(a)$.
✏️ Example 1 — The equation of a tangent line
Find the equation of the line tangent to the graph of $f(x)=\dfrac1x$ at $x=2$.
Step 1 The point: $f(2)=\tfrac12$, so the tangent passes through $\left(2,\tfrac12\right)$.
Step 2 The slope: $f(x)=x^{-1}$, so $f'(x)=-x^{-2}=-\dfrac{1}{x^2}$ and $f'(2)=-\dfrac14$.
Step 3 Assemble the equation: $$y=f(2)+f'(2)(x-2)=\frac12-\frac14(x-2)=-\frac14x+1.$$
Check The limit definition gives the same slope: $$\lim_{h\to0}\frac{\dfrac{1}{2+h}-\dfrac12}{h}=\lim_{h\to0}\frac{-h}{2h(2+h)}=\lim_{h\to0}\frac{-1}{2(2+h)}=-\frac14.$$ The differentiation rules are simply a faster route to the same number.
✏️ Example 2 — Finding the equation of a tangent line
Find the equation of the line tangent to the graph of $f(x)=x^2-4x+6$ at $x=1$.
Step 1 The point: $f(1)=1-4+6=3$, so the tangent passes through $(1,3)$.
Step 2 The slope: $f'(x)=2x-4$, so $f'(1)=-2$.
Step 3 Point-slope form: $$y=-2(x-1)+3=-2x+5.$$
✏️ Example 3 — Tangent lines needing the trig and chain rules
Find the equation of the line tangent to the graph of
  • a $f(x)=\cot x$ at $x=\dfrac{\pi}{4}$;
  • b $g(x)=\sqrt{x^2+9}$ at $x=4$.
(a) Step 1 The point: $f\!\left(\tfrac{\pi}{4}\right)=\cot\tfrac{\pi}{4}=1$, so the tangent passes through $\left(\tfrac{\pi}{4},1\right)$.
(a) Step 2 The slope: $f'(x)=-\csc^2x$, and $\csc\tfrac{\pi}{4}=\sqrt2$, so $f'\!\left(\tfrac{\pi}{4}\right)=-\left(\sqrt2\right)^2=-2$.
(a) Step 3 Point-slope form: $$y=-2\left(x-\frac{\pi}{4}\right)+1=-2x+\frac{\pi}{2}+1.$$
(b) Step 1 The point: $g(4)=\sqrt{16+9}=5$, so the tangent passes through $(4,5)$.
(b) Step 2 The slope, by the chain rule: $$g'(x)=\frac{1}{2\sqrt{x^2+9}}\cdot2x=\frac{x}{\sqrt{x^2+9}},\qquad g'(4)=\frac45.$$
(b) Step 3 Point-slope form: $$y=\frac45(x-4)+5=\frac45x+\frac95.$$

The Normal Line

📘 Definition — Normal line
The normal line to the graph of $f$ at $(a,f(a))$ is the line through that point perpendicular to the tangent line. Perpendicular lines have slopes whose product is $-1$, so if $f'(a)\neq0$, $$m_{\text{normal}}=-\frac{1}{f'(a)},\qquad y=f(a)-\frac{1}{f'(a)}(x-a).$$ Special cases. If $f'(a)=0$ the tangent is horizontal, $y=f(a)$, and the normal is the vertical line $x=a$. Conversely, where a curve has a vertical tangent $x=a$, the normal is the horizontal line $y=f(a)$.
⚠️ Negative reciprocal, not just negative
A common slip is to take the normal slope as $-f'(a)$ or $\dfrac{1}{f'(a)}$. It is the negative reciprocal: slope $2$ gives normal slope $-\tfrac12$; slope $-\tfrac34$ gives normal slope $\tfrac43$. Also note that a normal line only looks perpendicular on a graph whose two axes use the same scale.
✏️ Example 4 — Tangent and normal lines
For $f(x)=x^3-3x+1$, find the equations of the tangent line and the normal line
  • a at $x=2$;
  • b at $x=1$.
Step 1 $f'(x)=3x^2-3$.
(a) Tangent $f(2)=8-6+1=3$ and $f'(2)=12-3=9$, so $$y=9(x-2)+3=9x-15.$$
(a) Normal The normal slope is $-\tfrac19$: $$y=-\frac19(x-2)+3=-\frac19x+\frac{29}{9}.$$
(b) $f(1)=1-3+1=-1$ and $f'(1)=0$. The tangent is horizontal, $y=-1$, so the normal is the vertical line $x=1$. (The formula $-1/f'(a)$ cannot be used here — this is the special case.)
Interactive 1 — Tangent & Normal Line Explorer

A point slides back and forth along the curve carrying its tangent line (blue) and its normal line (purple, dashed). The readouts give both equations at every instant. The orange ring marks the point used in the example; press the jump button to stop there. Where possible the axes use equal scales, so the normal really looks perpendicular.

$x = a$—
$f(a)$—
slope $f'(a)$—
tangent line—
normal line—

Horizontal, Parallel and Perpendicular Tangents

📐 Tangent lines with a prescribed direction
Because the slope of the tangent at $x$ is $f'(x)$, a condition on the direction of the tangent becomes an equation for $x$:
Horizontal tangent$f'(x)=0$
Parallel to $y=mx+b$$f'(x)=m$
Perpendicular to $y=mx+b$ ($m\neq0$)$f'(x)=-\dfrac1m$
Solve the equation for $x$, then use $f(x)$ to find the point(s) on the curve.
✏️ Example 5 — Where a function has a horizontal tangent
Determine the values of $x$ for which $f(x)=x^3-7x^2+8x+1$ has a horizontal tangent line.
Step 1 A horizontal tangent means slope zero, so solve $f'(x)=0$.
Step 2 Differentiate and factor: $$f'(x)=3x^2-14x+8=(3x-2)(x-4).$$
Step 3 Solve $(3x-2)(x-4)=0$: the function has horizontal tangent lines at $x=\dfrac23$ and $x=4$.
Step 4 The points of tangency are $\left(\tfrac23,f\!\left(\tfrac23\right)\right)=\left(\tfrac23,\tfrac{95}{27}\right)$ and $(4,f(4))=(4,-15)$, so the horizontal tangents are $y=\tfrac{95}{27}$ and $y=-15$.
🧭 Problem-Solving Strategy — Tangent and normal lines
  1. Find the point. If only $x=a$ is given, compute $f(a)$. For an implicit curve, check that the given point satisfies the equation (or solve for the missing coordinate).
  2. Find the slope. Differentiate — with the rules, implicitly, or parametrically — and evaluate at the point: $m=f'(a)$.
  3. Write the tangent in point-slope form $y-y_0=m(x-x_0)$, then simplify if asked.
  4. Write the normal with slope $-\dfrac1m$. If $m=0$ the normal is $x=x_0$; if the tangent is vertical the normal is $y=y_0$.
  5. Working backwards? If the problem prescribes a direction or a point the line must pass through, leave the point of tangency as an unknown $a$, write the line in terms of $a$, and solve for $a$.

Tangent Lines to Implicit and Parametric Curves

Many important curves — circles, ellipses, the paths of moving objects — are not graphs of a single function $y=f(x)$. Their tangent and normal lines are found in exactly the same way; only the method for the slope $\dfrac{dy}{dx}$ changes.

🔁 Recap — Implicit curves (see 2.4)
For a curve $F(x,y)=0$, differentiate both sides with respect to $x$, treating $y$ as a function of $x$ (so $\dfrac{d}{dx}\big(y^n\big)=ny^{n-1}\dfrac{dy}{dx}$), and solve for $\dfrac{dy}{dx}$. The slope of the tangent at a point $(x_0,y_0)$ of the curve is the value of $\dfrac{dy}{dx}$ there, and the tangent is $y-y_0=\left.\dfrac{dy}{dx}\right|_{(x_0,y_0)}(x-x_0)$.
🔁 Recap — Parametric curves (see 2.5)
For a curve $x=x(t)$, $y=y(t)$, the slope of the tangent at the point with parameter $t$ is $$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\qquad\left(\frac{dx}{dt}\neq0\right).$$ The tangent is horizontal where $\dfrac{dy}{dt}=0$ and $\dfrac{dx}{dt}\neq0$, and vertical where $\dfrac{dx}{dt}=0$ and $\dfrac{dy}{dt}\neq0$. The point of tangency is $\big(x(t),y(t)\big)$.
✏️ Example 6 — Tangent lines found implicitly
Find the equation of the line tangent to
  • a the circle $x^2+y^2=25$ at the point $(3,-4)$;
  • b the curve $y^3+x^3-3xy=0$ at the point $\left(\tfrac32,\tfrac32\right)$.
(a) Step 1 Differentiate implicitly: $2x+2y\dfrac{dy}{dx}=0$, so $\dfrac{dy}{dx}=-\dfrac{x}{y}$. At $(3,-4)$: $$\left.\frac{dy}{dx}\right|_{(3,-4)}=-\frac{3}{-4}=\frac34.$$
(a) Step 2 Point-slope form: $$y=\frac34(x-3)+(-4)\quad\text{or}\quad y=\frac34x-\frac{25}{4}.$$ (The normal, slope $-\tfrac43$, is $y=-\tfrac43x$ — it passes through the centre of the circle, as every normal to a circle does.)
(b) Step 1 Differentiate both sides (product rule on $3xy$): $$3y^2\frac{dy}{dx}+3x^2-\left(3y+3x\frac{dy}{dx}\right)=0.$$
(b) Step 2 Collect the $\dfrac{dy}{dx}$ terms: $$\frac{dy}{dx}\left(3y^2-3x\right)=3y-3x^2\ \Longrightarrow\ \frac{dy}{dx}=\frac{y-x^2}{y^2-x}.$$
(b) Step 3 Substitute $\left(\tfrac32,\tfrac32\right)$: $$\left.\frac{dy}{dx}\right|=\frac{\tfrac32-\tfrac94}{\tfrac94-\tfrac32}=\frac{-\tfrac34}{\tfrac34}=-1.$$
(b) Step 4 Point-slope form: $y=-1\left(x-\tfrac32\right)+\tfrac32$, that is $y=-x+3$.
✏️ Example 7 — Tangent line to a parametric curve
Find the equation of the tangent line to the curve $x(t)=t^2-3$, $y(t)=2t-1$, $-3\le t\le4$, when $t=2$.
Step 1 $\dfrac{dx}{dt}=2t$ and $\dfrac{dy}{dt}=2$, so $\dfrac{dy}{dx}=\dfrac{2}{2t}=\dfrac1t$, and at $t=2$ the slope is $\dfrac12$.
Step 2 The point: $x(2)=4-3=1$ and $y(2)=4-1=3$, so the tangent passes through $(1,3)$.
Step 3 Point-slope form: $$y-3=\frac12(x-1)\ \Longrightarrow\ y=\frac12x+\frac52.$$
Note At $t=0$ we have $\dfrac{dx}{dt}=0$ but $\dfrac{dy}{dt}=2\neq0$: the curve has a vertical tangent $x=-3$ at the point $(-3,-1)$.
Interactive 2 — Tangents on Implicit and Parametric Curves

Implicit curves are drawn straight from their equation $F(x,y)=0$; parametric curves are traced from $\big(x(t),y(t)\big)$. Either way a point travels along the curve carrying its tangent (blue) and normal (purple, dashed). Watch for vertical tangents, where $\dfrac{dy}{dx}$ is undefined and the normal becomes horizontal.

point $(x, y)$—
parameter $t$—
slope $dy/dx$—
tangent line—
normal line—

Application — The Tangent Line as an Approximation

Zoom in far enough on a differentiable curve and it becomes indistinguishable from its tangent line. This is why the tangent line is used to estimate values of a function near a point where the function is easy to evaluate, and why an object that stops being constrained to a curved path (a released sling stone, a skidding car) moves off along the tangent.

📘 Definition — Linear approximation
The tangent line at $a$, written as a function, $$L(x)=f(a)+f'(a)(x-a),$$ is called the linear approximation (or linearisation) of $f$ at $a$. For $x$ close to $a$, $f(x)\approx L(x)$.
✏️ Example 8 — Estimating a square root with a tangent line
Use the tangent line to $f(x)=\sqrt{x}$ at $x=4$ to estimate $\sqrt{4.1}$, and compare the estimate with the true value.
Step 1 The point: $f(4)=2$. The slope: $f'(x)=\dfrac{1}{2\sqrt x}$, so $f'(4)=\dfrac14$.
Step 2 The tangent line is $$L(x)=2+\frac14(x-4)=\frac14x+1.$$
Step 3 Estimate: $\sqrt{4.1}\approx L(4.1)=2+\dfrac14(0.1)=2.025$.
Step 4 A calculator gives $\sqrt{4.1}=2.024845\ldots$, so the error is about $0.000154$ — less than $0.01\%$. The estimate is slightly too large because the graph of $\sqrt x$ bends below its tangent line.

Exercises

I–VI. Equations of tangent lines

I
Find the slope of the tangent line at $x=a$ and the equation of the tangent line to $f$ at $x=a$. (These solutions work from the limit definition of the derivative; check each slope with the differentiation rules.)
  • 1 $f(x)=3-4x$,  $a=2$
  • 2 $f(x)=x^2+x$,  $a=2$
  • 3 $f(x)=1-x-x^2$,  $a=0$
  • 4 $f(x)=2-3x^2$,  $a=-2$
  • 5 $f(x)=x^5+6$,  $a=-1$
Secant closing in on the tangent
increment $h$—
secant slope—
tangent slope—
tangent line—
1 $\dfrac{f(2+h)-f(2)}{h}=\dfrac{\left[3-4(2+h)\right]-(-5)}{h}=\dfrac{-4h}{h}=-4$, so $m=-4$. With $f(2)=-5$: $y=-4(x-2)-5=-4x+3$. (The tangent to a line is the line itself.)
2 $\dfrac{(2+h)^2+(2+h)-6}{h}=\dfrac{h^2+5h}{h}=h+5\to5$. With $f(2)=6$: $y=5(x-2)+6=5x-4$.
3 $\dfrac{\left[1-h-h^2\right]-1}{h}=-1-h\to-1$. With $f(0)=1$: $y=-x+1$.
4 $\dfrac{\left[2-3(-2+h)^2\right]-(-10)}{h}=\dfrac{12h-3h^2}{h}=12-3h\to12$. With $f(-2)=-10$: $y=12(x+2)-10=12x+14$.
5 $\dfrac{(-1+h)^5+6-5}{h}$. Expanding, $(-1+h)^5=-1+5h-10h^2+10h^3-5h^4+h^5$, so the quotient is $5-10h+10h^2-5h^3+h^4\to5$. With $f(-1)=5$: $y=5(x+1)+5=5x+10$.
I
Continued.
  • 6 $f(x)=\dfrac{-3}{x-1}$,  $a=4$
  • 7 $f(x)=\dfrac{2}{x+3}$,  $a=-4$
  • 8 $f(x)=\dfrac{3}{x^2}$,  $a=3$
  • 9 $f(x)=\sqrt{x+8}$,  $a=1$
Secant closing in on the tangent
increment $h$—
secant slope—
tangent slope—
tangent line—
6 $f(4)=-1$. The quotient is $$\frac{\dfrac{-3}{3+h}+1}{h}=\frac{\dfrac{-3+3+h}{3+h}}{h}=\frac{1}{3+h}\longrightarrow\frac13.$$ So $m=\tfrac13$ and $y=\tfrac13(x-4)-1=\tfrac13x-\tfrac73$.
7 $f(-4)=-2$. The quotient is $$\frac{\dfrac{2}{-1+h}+2}{h}=\frac{\dfrac{2+2(-1+h)}{-1+h}}{h}=\frac{2}{-1+h}\longrightarrow-2.$$ So $m=-2$ and $y=-2(x+4)-2=-2x-10$.
8 $f(3)=\tfrac13$. The quotient is $$\frac{\dfrac{3}{(3+h)^2}-\dfrac13}{h}=\frac{9-(3+h)^2}{3h(3+h)^2}=\frac{-6h-h^2}{3h(3+h)^2}=\frac{-(6+h)}{3(3+h)^2}\longrightarrow\frac{-6}{27}=-\frac29.$$ So $m=-\tfrac29$ and $y=-\tfrac29(x-3)+\tfrac13=-\tfrac29x+1$.
9 $f(1)=3$. Multiply by the conjugate: $$\frac{\sqrt{9+h}-3}{h}=\frac{(9+h)-9}{h\left(\sqrt{9+h}+3\right)}=\frac{1}{\sqrt{9+h}+3}\longrightarrow\frac16.$$ So $m=\tfrac16$ and $y=\tfrac16(x-1)+3=\tfrac16x+\tfrac{17}{6}$.
II
Find the equation of the tangent line to the graph of the given function at the indicated point.
  • 1 $y=3x^2+4x+1$ at $(0,1)$
  • 2 $y=2\sqrt{x}+1$ at $(4,5)$
  • 3 $y=\dfrac{2x}{x-1}$ at $(-1,1)$
  • 4 $y=\dfrac2x-\dfrac{3}{x^2}$ at $(1,-1)$
Tangent line lab
increment $h$—
secant slope—
tangent slope—
tangent line—
1 $y'=6x+4$, so at $x=0$ the slope is $4$: $y=4x+1$.
2 $y=2x^{1/2}+1$, so $y'=\dfrac{1}{\sqrt{x}}$ and at $x=4$ the slope is $\tfrac12$: $$y=\tfrac12(x-4)+5=\tfrac12x+3.$$
3 Quotient rule: $y'=\dfrac{2(x-1)-2x}{(x-1)^2}=\dfrac{-2}{(x-1)^2}$, so at $x=-1$ the slope is $-\tfrac12$: $$y=-\tfrac12(x+1)+1=-\tfrac12x+\tfrac12.$$
4 $y=2x^{-1}-3x^{-2}$, so $y'=-\dfrac{2}{x^2}+\dfrac{6}{x^3}$, which at $x=1$ is $4$: $$y=4(x-1)-1=4x-5.$$
III
III. Find the equation of the line tangent to the graph of $f(x)=3x^2-11$ at $x=2$.
IV. Find the equation of the line tangent to the graph of $f(x)=2x^3+4x^2-5x-3$ at $x=-1$.
V. Find the equation of the line tangent to the graph of $f(x)=x^2+\dfrac4x-10$ at $x=8$.
VI. Find the equation of the line tangent to the graph of $f(x)=\left(3x-x^2\right)\left(3-x-x^2\right)$ at $x=1$.
Tangent line lab
increment $h$—
secant slope—
tangent slope—
tangent line—
III $f'(x)=6x$, so $f'(2)=12$; $f(2)=12-11=1$. Tangent: $y=12(x-2)+1=12x-23$.
IV $f'(x)=6x^2+8x-5$, so $f'(-1)=6-8-5=-7$; $f(-1)=-2+4+5-3=4$. Tangent: $y=-7(x+1)+4=-7x-3$.
V $f'(x)=2x-\dfrac{4}{x^2}$, so $f'(8)=16-\dfrac{4}{64}=\dfrac{255}{16}$; $f(8)=64+\tfrac12-10=\dfrac{109}{2}$. Tangent: $$y=\frac{255}{16}(x-8)+\frac{109}{2}=\frac{255}{16}x-73.$$
VI Product rule: $$f'(x)=(3-2x)\left(3-x-x^2\right)+\left(3x-x^2\right)(-1-2x).$$ At $x=1$: $f'(1)=(1)(1)+(2)(-3)=-5$, and $f(1)=(2)(1)=2$. Tangent: $y=-5(x-1)+2=-5x+7$.

VII–X. Tangent lines with trigonometric, exponential and composite functions

VII
VII. Find an equation of the tangent line to the graph of $f(x)=4xe^{x^2-1}$ at the point where $x=-1$.
VIII. Find the equation of the line tangent to the graph of $y=\log_2(3x+1)$ at $x=1$.
IX. Find the equation of the tangent line to the graph of $h(x)=\dfrac{1}{(3x-5)^2}$ at $x=2$.
X. Find the equation of the tangent line to $y=-\sin\left(\dfrac{x}{2}\right)$ at the origin.
Tangent line lab
increment $h$—
secant slope—
tangent slope—
tangent line—
VII By the product and chain rules, $f'(x)=4e^{x^2-1}+4x\cdot2xe^{x^2-1}=\left(8x^2+4\right)e^{x^2-1}$, so $f'(-1)=12e^0=12$, and $f(-1)=-4e^0=-4$. Tangent: $$y=12(x+1)-4=12x+8.$$
VIII $y'=\dfrac{3}{(3x+1)\ln2}$, so at $x=1$ the slope is $\dfrac{3}{4\ln2}\approx1.0820$, and $y(1)=\log_24=2$. Tangent: $$y=\frac{3}{4\ln2}(x-1)+2.$$
IX $h(x)=(3x-5)^{-2}$, so $h'(x)=-2(3x-5)^{-3}\cdot3=\dfrac{-6}{(3x-5)^3}$. At $x=2$: $h'(2)=-6$ and $h(2)=1$. Tangent: $$y=-6(x-2)+1=-6x+13.$$
X $y'=-\dfrac12\cos\left(\dfrac{x}{2}\right)$, so at $x=0$ the slope is $-\dfrac12$, and the curve passes through the origin. Tangent: $$y=-\frac{x}{2}.$$

XI–XIV. Horizontal and parallel tangents — working backwards from a condition

XI
Find the $x$-coordinates at which the tangent line to $y=\left(x-\dfrac6x\right)^8$ is horizontal.
Where does $f'$ cross zero?
$f$ with its tangent
$f'$ being traced
$x$—
$f(x)$—
$f'(x)$—
$f$ is…—
Step 1 Differentiate with the power rule for compositions: $$\frac{dy}{dx}=8\left(x-\frac6x\right)^7\left(1+\frac{6}{x^2}\right).$$
Step 2 The second factor $1+\dfrac{6}{x^2}$ is always positive, so the derivative vanishes only when $$x-\frac6x=0\ \Longrightarrow\ x^2=6\ \Longrightarrow\ x=\pm\sqrt6.$$
Step 3 So the tangent line is horizontal at $x=\sqrt6\approx2.449$ and $x=-\sqrt6\approx-2.449$. At both points $y=0$, since the bracket itself is zero.
XII
XII. Find the point on the graph of $f(x)=x^3$ such that the tangent line at that point has an $x$-intercept of $6$.
XIII. Find the equation of the line passing through the point $(3,3)$ and tangent to the graph of $f(x)=\dfrac{6}{x-1}$.
XIV. Find the values of $x$ for which the line tangent to the graph of $f(x)=4x^2-3x+2$ is parallel to the line $y=2x+3$.
Slide the point of tangency until the condition is met
$a$—
the condition asks for—
currently—
tangent line—
XII At the point $\left(a,a^3\right)$ the slope is $3a^2$, so the tangent is $y=3a^2(x-a)+a^3$. Setting $y=0$ and solving for $x$: $$x=a-\frac{a^3}{3a^2}=a-\frac{a}{3}=\frac{2a}{3}.$$ We need $\dfrac{2a}{3}=6$, so $a=9$ and the point is $\left(9,729\right)$.
XIII At $\left(a,\dfrac{6}{a-1}\right)$ the slope is $-\dfrac{6}{(a-1)^2}$. Requiring the tangent to pass through $(3,3)$: $$3=-\frac{6(3-a)}{(a-1)^2}+\frac{6}{a-1}.$$ Multiplying by $(a-1)^2$ gives $3(a-1)^2=6a-18+6a-6$, that is $3a^2-18a+27=0$, so $(a-3)^2=0$ and $a=3$. The slope is $-\tfrac64=-\tfrac32$ and the line is $$y=-\tfrac32(x-3)+3=-\tfrac32x+\tfrac{15}{2}.$$ Note that $(3,3)$ is itself a point of the curve, so the tangent there is the answer.
XIV Parallel lines have equal slopes, so solve $f'(x)=2$: $$8x-3=2\ \Longrightarrow\ x=\frac58.$$

XV–XVII. Tangent lines to implicit curves

XV
Find the equation of the line tangent to the hyperbola $x^2-y^2=16$ at the point $(5,3)$.
Trace the hyperbola
$x$—
$y$—
$dy/dx$—
tangent line—
Step 1 $2x-2y\,y^\prime=0\ \Rightarrow\ \dfrac{dy}{dx}=\dfrac{x}{y}$, so the slope at $(5,3)$ is $\dfrac53$.
Step 2 $y=\dfrac53(x-5)+3=\dfrac53x-\dfrac{16}{3}$.
XVI
Find the equation of the tangent line to the graph of the given equation at the indicated point.
  • 1 $x^4y-xy^3=-2$,  $(-1,-1)$
  • 2 $x^2y^2+5xy=14$,  $(2,1)$
  • 3 $\tan(xy)=y$,  $\left(\tfrac{\pi}{4},1\right)$
  • 4 $xy^2+\sin(\pi y)-2x^2=10$,  $(2,-3)$
  • 5 $xy+\sin x=1$,  $\left(\tfrac{\pi}{2},0\right)$
  • 6 $\dfrac{x}{y}+5x-7=-\dfrac34y$,  $(1,2)$
Trace each curve to the given point
$x$—
$y$—
$dy/dx$—
tangent line—
1 $4x^3y+x^4y^\prime-y^3-3xy^2y^\prime=0\ \Rightarrow\ y^\prime=\dfrac{y^3-4x^3y}{x^4-3xy^2}$. At $(-1,-1)$: $y^\prime=\dfrac{-1-4}{1+3}=-\dfrac54$. Tangent: $y=-\dfrac54(x+1)-1=-\dfrac54x-\dfrac94$.
2 $2xy^2+2x^2yy^\prime+5y+5xy^\prime=0\ \Rightarrow\ y^\prime=-\dfrac{2xy^2+5y}{2x^2y+5x}$. At $(2,1)$: $y^\prime=-\dfrac{4+5}{8+10}=-\dfrac12$. Tangent: $y=-\dfrac12(x-2)+1=-\dfrac12x+2$.
3 $\sec^2(xy)\left(y+xy^\prime\right)=y^\prime$. At $\left(\tfrac\pi4,1\right)$, $\sec^2\tfrac\pi4=2$: $2\left(1+\tfrac\pi4y^\prime\right)=y^\prime\Rightarrow y^\prime\left(1-\tfrac\pi2\right)=2\Rightarrow y^\prime=\dfrac{4}{2-\pi}\approx-3.504$. Tangent: $y=\dfrac{4}{2-\pi}\left(x-\dfrac\pi4\right)+1$.
4 $y^2+2xyy^\prime+\pi\cos(\pi y)y^\prime-4x=0$. At $(2,-3)$: $9-12y^\prime-\pi y^\prime-8=0\Rightarrow y^\prime=\dfrac{1}{12+\pi}\approx0.0660$. Tangent: $y=\dfrac{1}{12+\pi}(x-2)-3$.
5 $y+xy^\prime+\cos x=0\Rightarrow y^\prime=-\dfrac{y+\cos x}{x}$. At $\left(\tfrac\pi2,0\right)$: $y^\prime=-\dfrac{0+0}{\pi/2}=0$. Tangent: $y=0$.
6 $\dfrac{y-xy^\prime}{y^2}+5=-\dfrac34y^\prime$. At $(1,2)$: $\dfrac{2-y^\prime}{4}+5=-\dfrac34y^\prime\Rightarrow 2-y^\prime+20=-3y^\prime\Rightarrow y^\prime=-11$. Tangent: $y=-11(x-1)+2=-11x+13$.
XVII
Find the equation of the tangent line to the graph of $x^3-x\ln y+y^3=2x+5$ at the point where $x=2$.
Trace the curve
$x$—
$y$—
$dy/dx$—
tangent line—
Step 1 At $x=2$: $8-2\ln y+y^3=9$, i.e. $y^3-2\ln y-1=0$, which is satisfied by $y=1$ (since $\ln1=0$). So the point is $(2,1)$.
Step 2 Differentiate: $3x^2-\ln y-\dfrac{x}{y}y^\prime+3y^2y^\prime=2$. At $(2,1)$: $12-0-2y^\prime+3y^\prime=2\Rightarrow y^\prime=-10$.
Step 3 Tangent: $y=-10(x-2)+1=-10x+21$.

XVIII–XIX. Normal lines

XVIII
XVIII. For the equation $x^2+2xy-3y^2=0$, find the equation of the normal to the tangent line at the point $(1,1)$.
XIX. For the equation $2x^3+2y^3-9xy=0$, find the equation of the normal to the tangent line at the point $(2,1)$.
Tangent (blue) and normal (purple, dashed)
$x$—
$y$—
$dy/dx$—
tangent line—
XVIII $2x+2y+2xy^\prime-6yy^\prime=0\Rightarrow y^\prime=\dfrac{x+y}{3y-x}$. At $(1,1)$ the tangent slope is $\dfrac{2}{2}=1$, so the normal slope is $-1$: $$y=-(x-1)+1=-x+2.$$ (The curve factors as $(x+3y)(x-y)=0$ — two lines; at $(1,1)$ we are on $y=x$.)
XIX $6x^2+6y^2y^\prime-9y-9xy^\prime=0\Rightarrow y^\prime=\dfrac{9y-6x^2}{6y^2-9x}=\dfrac{3y-2x^2}{2y^2-3x}$. At $(2,1)$: $y^\prime=\dfrac{3-8}{2-6}=\dfrac54$, so the normal slope is $-\dfrac45$: $$y=-\frac45(x-2)+1=-\frac45x+\frac{13}{5}.$$

XX. Tangent lines to parametric curves

XX
Determine the slope of the tangent line, then find the equation of the tangent line at the given value of the parameter.
  • 1 $x=3\sin t$, $y=3\cos t$,  $t=\tfrac{\pi}{4}$
  • 2 $x=\cos t$, $y=8\sin t$,  $t=\tfrac{\pi}{2}$
  • 3 $x=2t$, $y=t^3$,  $t=-1$
  • 4 $x=t+\dfrac1t$, $y=t-\dfrac1t$,  $t=1$
  • 5 $x=\sqrt t$, $y=2t$,  $t=4$
Jump to the given t
$t$—
$(x, y)$—
rates—
$dy/dx$—
tangent line—
1 $\dfrac{dy}{dx}=\dfrac{-3\sin t}{3\cos t}=-\tan t$, so the slope is $-1$. Point: $\left(\tfrac{3\sqrt2}{2},\tfrac{3\sqrt2}{2}\right)$. Tangent: $y=-x+3\sqrt2$.
2 $\dfrac{dy}{dx}=\dfrac{8\cos t}{-\sin t}$, which is $0$ at $t=\tfrac\pi2$. Point: $(0,8)$. Tangent: $y=8$.
3 $\dfrac{dy}{dx}=\dfrac{3t^2}{2}$, so the slope is $\tfrac32$. Point: $(-2,-1)$. Tangent: $y=\tfrac32(x+2)-1=\tfrac32x+2$.
4 $x^\prime=1-\dfrac{1}{t^2}$ and $y^\prime=1+\dfrac{1}{t^2}$. At $t=1$, $x^\prime=0$ while $y^\prime=2$: the tangent is vertical. Point: $(2,0)$. Tangent: $x=2$.
5 $\dfrac{dy}{dx}=\dfrac{2}{1/(2\sqrt t)}=4\sqrt t$, so the slope is $8$. Point: $(2,8)$. Tangent: $y=8(x-2)+8=8x-8$.

XXI–XXIII. More practice — tangent and normal lines, approximation and applications

XXI
Find the equation of the tangent line and the equation of the normal line to the graph of $f$ at the given value of $x$.
  • 1 $f(x)=x^2-2x-3$,  $x=3$
  • 2 $f(x)=\sqrt{2x+1}$,  $x=4$
  • 3 $f(x)=e^{2x}$,  $x=0$
  • 4 $f(x)=x\ln x$,  $x=e$
  • 5 $f(x)=\sin 2x$,  $x=\dfrac{\pi}{4}$
  • 6 $f(x)=\dfrac{x+1}{x-1}$,  $x=2$
Tangent and normal at the given point
$x = a$—
$f(a)$—
slope $f'(a)$—
tangent line—
normal line—
1 $f(3)=9-6-3=0$ and $f'(x)=2x-2$, so $f'(3)=4$. Tangent: $y=4(x-3)=4x-12$. Normal (slope $-\tfrac14$): $$y=-\frac14(x-3)=-\frac14x+\frac34.$$
2 $f(4)=\sqrt9=3$ and, by the chain rule, $f'(x)=\dfrac{1}{2\sqrt{2x+1}}\cdot2=\dfrac{1}{\sqrt{2x+1}}$, so $f'(4)=\tfrac13$. Tangent: $y=\tfrac13(x-4)+3=\tfrac13x+\tfrac53$. Normal (slope $-3$): $y=-3(x-4)+3=-3x+15$.
3 $f(0)=1$ and $f'(x)=2e^{2x}$, so $f'(0)=2$. Tangent: $y=2x+1$. Normal (slope $-\tfrac12$): $y=-\tfrac12x+1$.
4 $f(e)=e\ln e=e$. Product rule: $f'(x)=\ln x+x\cdot\dfrac1x=\ln x+1$, so $f'(e)=2$. Tangent: $y=2(x-e)+e=2x-e$. Normal (slope $-\tfrac12$): $$y=-\frac12(x-e)+e=-\frac12x+\frac{3e}{2}.$$
5 $f\!\left(\tfrac\pi4\right)=\sin\tfrac\pi2=1$ and $f'(x)=2\cos2x$, so $f'\!\left(\tfrac\pi4\right)=2\cos\tfrac\pi2=0$. The tangent is horizontal, $y=1$, so the normal is the vertical line $x=\dfrac{\pi}{4}$.
6 $f(2)=3$. Quotient rule: $f'(x)=\dfrac{(x-1)-(x+1)}{(x-1)^2}=\dfrac{-2}{(x-1)^2}$, so $f'(2)=-2$. Tangent: $y=-2(x-2)+3=-2x+7$. Normal (slope $\tfrac12$): $y=\tfrac12(x-2)+3=\tfrac12x+2$.
XXII
Applications of the tangent line.
  • 1 Use the tangent line to $f(x)=\sqrt[3]{x}$ at $x=8$ to estimate $\sqrt[3]{8.2}$. Compare with the calculator value.
  • 2 Use the tangent line to $f(x)=\ln x$ at $x=1$ to estimate $\ln1.05$. Is the estimate too large or too small?
  • 3 Headlights. A car travels along the curved road $y=\dfrac{x^2}{50}$ (in metres), moving to the right. Its headlights point along the tangent to the road. When the car is at the point $(10,2)$, does the beam hit a road sign at $(30,10)$?
  • 4 Sensor linearisation. The resistance of a thermistor is $R(T)=100e^{-0.04(T-25)}$ ohms at temperature $T\ ^\circ$C. Find the linear approximation of $R$ at the operating point $T=25$, use it to estimate $R(27)$, and compare with the exact value.
  • 5 A sling. A stone is whirled counterclockwise around the circle $x^2+y^2=1$ (in metres) and released at the point $\left(\tfrac35,\tfrac45\right)$. After release it travels along the tangent line. Find the equation of that line and the point where the stone hits the wall $x=-2$. (Ignore gravity.)
The tangent line as an approximation
$x = a$—
$f(a)$—
slope $f'(a)$—
tangent line—
normal line—
1 $f(8)=2$ and $f'(x)=\tfrac13x^{-2/3}$, so $f'(8)=\dfrac{1}{3\cdot4}=\dfrac1{12}$. Tangent: $L(x)=2+\tfrac1{12}(x-8)$. Then $$\sqrt[3]{8.2}\approx L(8.2)=2+\frac{0.2}{12}=2.01\overline{6}\approx2.0167.$$ The calculator gives $2.016530$, so the error is about $0.00014$.
2 $f(1)=0$ and $f'(x)=\dfrac1x$, so $f'(1)=1$ and the tangent is $L(x)=x-1$. Hence $\ln1.05\approx0.05$. The true value is $0.048790$, so the estimate is too large (by about $0.0012$): the graph of $\ln x$ bends below its tangent line.
3 $y'=\dfrac{2x}{50}=\dfrac{x}{25}$, so at $x=10$ the slope is $0.4$. The beam lies along $y=0.4(x-10)+2=0.4x-2$. At $x=30$ this gives $y=12-2=10$ — exactly the height of the sign. Yes, the beam hits the sign.
4 $R(25)=100$ and $R'(T)=100e^{-0.04(T-25)}\cdot(-0.04)=-4e^{-0.04(T-25)}$, so $R'(25)=-4$ ohms per degree. Linearisation: $L(T)=100-4(T-25)$, so $R(27)\approx100-8=92$ ohms. Exactly, $R(27)=100e^{-0.08}\approx92.31$ ohms: the linear model is off by about $0.31$ ohm (a $0.3\%$ error).
5 Implicitly, $\dfrac{dy}{dx}=-\dfrac{x}{y}$, so at $\left(\tfrac35,\tfrac45\right)$ the slope is $-\tfrac34$ and the tangent is $$y=-\frac34\left(x-\frac35\right)+\frac45=-\frac34x+\frac54.$$ Moving counterclockwise from this point the stone heads up and to the left, so it meets $x=-2$ at $y=-\tfrac34(-2)+\tfrac54=\tfrac{11}{4}$: the point $\left(-2,\tfrac{11}{4}\right)$. Note that the radius to the release point, $y=\tfrac43x$, is the normal line — the stone does not fly outward along it.
XXIII
Working backwards, and some concepts.
  • 1 Find the points on the graph of $y=x^3-3x$ at which the tangent line is parallel to the line $y=9x+1$.
  • 2 Find the point on the parabola $y=x^2$ at which the tangent line is perpendicular to the line $x+4y=1$, and write that tangent line.
  • 3 Find all points on the parabola $y=x^2$ whose normal line passes through the point $\left(0,\tfrac32\right)$, and write the normal lines.
  • 4 Find the equations of the two lines through the point $(1,-3)$ that are tangent to the parabola $y=x^2$.
  • 5 True or false? Explain.  (a) If $f'(a)=0$, the normal line at $a$ is $x=a$.  (b) Every normal line to a circle passes through its centre.  (c) A tangent line never crosses the curve at the point of tangency.
Slide the point of tangency until the condition is met
$a$—
the condition asks for—
currently—
line drawn—
1 Parallel means equal slopes: $y'=3x^2-3=9$, so $x^2=4$ and $x=\pm2$. The points are $(2,2)$ and $(-2,-2)$ (since $8-6=2$ and $-8+6=-2$).
2 The line $x+4y=1$ is $y=-\tfrac14x+\tfrac14$, with slope $-\tfrac14$. A perpendicular tangent needs slope $4$: $y'=2x=4$ gives $x=2$, so the point is $(2,4)$ and the tangent is $y=4(x-2)+4=4x-4$.
3 At $\left(a,a^2\right)$ with $a\neq0$ the normal slope is $-\dfrac1{2a}$ and the normal is $y=a^2-\dfrac{1}{2a}(x-a)$. At $x=0$ its height is $a^2+\tfrac12$; setting this equal to $\tfrac32$ gives $a=\pm1$. At $a=0$ the tangent is horizontal and the normal is the $y$-axis $x=0$, which also passes through $\left(0,\tfrac32\right)$. So there are three points: $(0,0)$ with normal $x=0$; $(1,1)$ with normal $y=-\tfrac12x+\tfrac32$; $(-1,1)$ with normal $y=\tfrac12x+\tfrac32$.
4 The point $(1,-3)$ is not on the parabola, so let the point of tangency be $\left(a,a^2\right)$. The tangent there is $y=2a(x-a)+a^2=2ax-a^2$. It passes through $(1,-3)$ when $-3=2a-a^2$, i.e. $a^2-2a-3=(a-3)(a+1)=0$. So $a=3$ or $a=-1$, and the tangent lines are $$y=6x-9\qquad\text{and}\qquad y=-2x-1.$$
5 (a) True: the tangent is horizontal, $y=f(a)$, so the perpendicular through $(a,f(a))$ is vertical, $x=a$. (b) True: for $x^2+y^2=r^2$ the tangent slope at $(x_0,y_0)$ is $-\tfrac{x_0}{y_0}$, so the normal slope is $\tfrac{y_0}{x_0}$ and the normal $y=\tfrac{y_0}{x_0}x$ passes through $(0,0)$ (when $x_0=0$ the normal is $x=0$, which also does). (c) False: the tangent to $y=x^3$ at $(0,0)$ is $y=0$, and the curve crosses it there, going from below to above.
🔬 Research–Teaching NexusResearch-oriented · optional, about 30 minutes

Research question. Over which interval is the tangent-line estimate of $\sqrt{x}$ at $x=4$ within 1% of the true value?

Why it matters. Engineers and physicists linearise hard functions every day (e.g. in control systems and circuit design); knowing where the approximation is accurate is essential.

What to do.

  1. Find the tangent line to $y=\sqrt{x}$ at $x=4$: $L(x)=2+\frac14(x-4)$.
  2. In a spreadsheet, compute the relative error $\frac{|\sqrt{x}-L(x)|}{\sqrt{x}}$ for $x$ from 1 to 9 in steps of 0.1.
  3. Find the interval where the error is at most 1%.
  4. Repeat for $\ln x$ at $x=1$ ($L(x)=x-1$) using the absolute error $|\ln x-(x-1)|\le0.01$.

Tools: Spreadsheet or Desmos

Hand in: A half-page note: the two intervals, a graph, and why the error grows away from the point.

Going further. Anywhere: test your own choice of function and point.

Instructor note: The $\sqrt{x}$ interval is about $3.02\le x\le5.31$ (not symmetric); for $\ln x$ it is about $0.864\le x\le1.148$.

🧪 Self-check quiz

0/5 correct

Q1. For $f(x)=\dfrac1x$, the tangent line at $x=2$ is:

Q2. The tangent to a curve at a point has slope $-\tfrac23$. The normal line there has slope:

Q3. $f(x)=x^3-7x^2+8x+1$ has horizontal tangents at:

Q4. Using the tangent line to $y=\sqrt x$ at $x=4$, the estimate of $\sqrt{4.1}$ is:

Q5. A stone whirled on a circle is released at a point $P$. Ignoring gravity, it moves off along:

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