Grade 12 Mathematics - Advanced Stream - Term 1: Limits and Continuity - Tangent Lines and Arc Length

Unit-2: Limits and Continuity - Lesson 2.1: A Brief Preview of Calculus: Tangent Lines and the Length of a Curve

Learning Objectives

• Estimate the slope for a given function at a given point using tables.

• Estimate an arc length for a given function.

Keywords

• Slope of the Curve: Represents the instantaneous rate of change of a function at a specific point, which is the slope of the tangent line to the curve at that point.

• Arc Length: The total distance along a curve between two points.

Introduction to Slope

• Slope of a Straight Line: For any two points $(x<em>1, y</em>1)$ and $(x<em>2, y</em>2)$ on a straight line, the slope is given by the formula:
  $Slope = \frac{\Delta y}{\Delta x} = \frac{y<em>2 - y</em>1}{x<em>2 - x</em>1}$
  This value remains constant for any two points on the same straight line.

• Slope of the Secant Line: A secant line connects two distinct points on a curve. Its slope is calculated using the same formula as a straight line:
  $Slope = \frac{\Delta y}{\Delta x} = \frac{y<em>2 - y</em>1}{x<em>2 - x</em>1}$
  The slope of a secant line represents the average rate of change of the function between the two points.

• Slope of the Curve at a Given Point (Tangent Line):

  • The slope of the tangent line to a function at a given point represents the instantaneous rate of change of the function at that specific point.

  • Calculus provides methods for approximating and ultimately finding this instantaneous change.

Estimating the Slope of a Curve

This method involves choosing a sequence of points near the given point and computing the slope of the secant line joining each of these points with the given point. As the chosen points get closer to the given point, the slope of the secant line approaches the slope of the tangent line (the instantaneous rate of change) at the given point.

Example 1: Estimate the slope of $y = x^2 + 1$ at $x = 1$ (Point $(1,2)$).

• For x > 1 (approaching from the right):

  • Second Point $(2,5)$: $m_{sec} = \frac{5-2}{2-1} = 3$

  • Second Point $(1.1,2.21)$: $m_{sec} = \frac{2.21-2}{1.1-1} = 2.1$

  • Second Point $(1.01,2.0201)$: $m_{sec} = \frac{2.0201-2}{1.01-1} = 2.01$

• For x < 1 (approaching from the left):

  • Second Point $(0,1)$: $m_{sec} = \frac{1-2}{0-1} = 1$

  • Second Point $(0.9,1.81)$: $m_{sec} = \frac{1.81-2}{0.9-1} = 1.9$

  • Second Point $(0.99,1.9801)$: $m_{sec} = \frac{1.9801-2}{0.99-1} = 1.99$

• Estimation: As the second point gets closer to $(1,2)$, the slope of the secant approaches $2$. Therefore, the estimated slope of the curve at $(1,2)$ is $2$.

Example 2: Estimate the slope of $y = \sin x$ at $x = 0$ (Point $(0,0)$).

• For x > 0 (approaching from the right):

  • Second Point $(1,\sin 1)$: $m_{sec} = \frac{\sin(1)-0}{1-0} \approx 0.84147$

  • Second Point $(0.1,\sin 0.1)$: $m_{sec} = \frac{\sin(0.1)-0}{0.1-0} \approx 0.99833$

  • Second Point $(0.01,\sin 0.01)$: $m_{sec} = \frac{\sin(0.01)-0}{0.01-0} \approx 0.99998$

• For x < 0 (approaching from the left):

  • Second Point $(-1,\sin(-1))$: $m_{sec} = \frac{\sin(-1)-0}{-1-0} \approx 0.84147$

  • Second Point $(-0.1,\sin(-0.1))$: $m_{sec} = \frac{\sin(-0.1)-0}{-0.1-0} \approx 0.99833$

  • Second Point $(-0.01,\sin(-0.01))$: $m_{sec} = \frac{\sin(-0.01)-0}{-0.01-0} \approx 0.99998$

• Estimation: As the second point gets closer to $(0,0)$, the slope of the secant approaches $1$. Therefore, the estimated slope of the curve at $(0,0)$ is $1$.

Example 3: Estimate the slope of $y = e^x$ at $x = 0$ (Point $(0,1)$).

• For x > 0 (approaching from the right):

  • Second Point $(1,e)$: $m_{sec} = \frac{e-1}{1-0} \approx 1.718282$

  • Second Point $(0.1,1.1052)$: $m_{sec} = \frac{1.1052-1}{0.1-0} \approx 1.051709$

  • Second Point $(0.01,1.0101)$: $m_{sec} = \frac{1.0101-1}{0.01-0} \approx 1.005017$

• For x < 0 (approaching from the left):

  • Second Point $(-1,0.3679)$: $m_{sec} = \frac{0.3679-1}{-1-0} \approx 0.632121$

  • Second Point $(-0.1,0.9048)$: $m_{sec} = \frac{0.9048-1}{-0.1-0} \approx 0.951626$

  • Second Point $(-0.01,0.9901)$: $m_{sec} = \frac{0.9901-1}{-0.01-0} \approx 0.995017$

• Estimation: As the second point gets closer to $(0,1)$, the slope of the secant approaches $1$. Therefore, the estimated slope of the curve at $(0,1)$ is $1$.

Exercise Example: Estimate the slope of $y = e^x$ at $x = 1$ (Point $(1,e)$).

• For x > 1 (approaching from the right):

  • Second Point $(2,7.3891)$: $m_{sec} = \frac{7.3891-e}{2-1} \approx 4.6708$

  • Second Point $(1.1,3.0042)$: $m_{sec} = \frac{3.0042-e}{1.1-1} \approx 2.859$

  • Second Point $(1.01,2.7456)$: $m_{sec} = \frac{2.7456-e}{1.01-1} \approx 2.73$

• For x < 1 (approaching from the left):

  • Second Point $(0,1)$: $m_{sec} = \frac{1-e}{0-1} \approx 1.71823$

  • Second Point $(0.9,2.4596)$: $m_{sec} = \frac{2.4596-e}{0.9-1} \approx 2.587$

  • Second Point $(0.99,2.6912)$: $m_{sec} = \frac{2.6912-e}{0.99-1} \approx 2.71$

• Estimation: As the second point gets closer to $(1,e)$, the slope of the secant approaches $e$\ (approximately $2.718$. Therefore, the estimated slope of the curve at $(1,e)$ is $e$.

Exercise Example: Estimate the slope of $y = \cos x$ at $x = 0$ (Point $(0,1)$).

• For x > 0 (approaching from the right):

  • Second Point $(1,\cos 1)$: $m_{sec} = \frac{\cos(1)-1}{1-0} \approx -0.4597$

  • Second Point $(0.1,\cos(0.1))$: $m_{sec} = \frac{\cos(0.1)-1}{0.1-0} \approx -0.04996$

  • Second Point $(0.01,\cos(0.01))$: $m_{sec} = \frac{\cos(0.01)-1}{0.01-0} \approx -0.005$

• For x < 0 (approaching from the left):

  • Second Point $(-1,\cos(-1))$: $m_{sec} = \frac{\cos(-1)-1}{-1-0} \approx 0.4597$

  • Second Point $(-0.1,\cos(-0.1))$: $m_{sec} = \frac{\cos(-0.1)-1}{-0.1-0} \approx 0.04996$

  • Second Point $(-0.01,\cos(-0.01))$: $m_{sec} = \frac{\cos(-0.01)-1}{-0.01-0} \approx 0.005$

• Estimation: As the second point gets closer to $(0,1)$, the slope of the secant approaches $0$. Therefore, the estimated slope of the curve at $(0,1)$ is $0$.

Exercise Example: Estimate the slope of $y = x^2 - 2x$ at $x = 2$ (Point $(2,0)$).

• For x > 2 (approaching from the right):

  • Second Point $(3,3)$: $m_{sec} = \frac{3-0}{3-2} = 3$

  • Second Point $(2.1,0.21)$: $m_{sec} = \frac{0.21-0}{2.1-2} = 2.1$

  • Second Point $(2.01,0.0201)$: $m_{sec} = \frac{0.0201-0}{2.01-2} = 2.01$

• For x < 2 (approaching from the left):

  • Second Point $(1,-1)$: $m_{sec} = \frac{-1-0}{1-2} = 1$

  • Second Point $(1.9,-0.19)$: $m_{sec} = \frac{-0.19-0}{1.9-2} = 1.9$

  • Second Point $(1.99,-0.0199)$: $m_{sec} = \frac{-0.0199-0}{1.99-2} = 1.99$

• Estimation: As the second point gets closer to $(2,0)$, the slope of the secant approaches $2$. Therefore, the estimated slope of the curve at $(2,0)$ is $2$.

Exercise Example: Estimate the slope of $y = \sin(2x)$ at $x = 0$ (Point $(0,0)$).

• For x > 0 (approaching from the right):

  • Second Point $(0.2,0.3894)$: $m_{sec} = \frac{0.3894-0}{0.2-0} \approx 1.9471$

  • Second Point $(0.1,0.1987)$: $m_{sec} = \frac{0.1987-0}{0.1-0} \approx 1.9876$

  • Second Point $(0.01,0.019999)$: $m_{sec} = \frac{0.019999-0}{0.01-0} \approx 1.9999$

• For x < 0 (approaching from the left):

  • Second Point $(-0.2,-0.3894)$: $m_{sec} = \frac{-0.3894-0}{-0.2-0} \approx 1.9471$

  • Second Point $(-0.1,-0.1987)$: $m_{sec} = \frac{-0.1987-0}{-0.1-0} \approx 1.9876$

  • Second Point $(-0.01,-0.02)$: $m_{sec} = \frac{-0.02-0}{-0.01-0} = 2$

• Estimation: As the second point gets closer to $(0,0)$, the slope of the secant approaches $2$. Therefore, the estimated slope of the curve at $(0,0)$ is $2$.

Estimating Arc Length of a Curve

• The arc length of a function over a given interval can be approximated by dividing the curve into n line segments.

• This is done by choosing points with evenly spaced $x$-coordinates along the interval.

• The distance between consecutive points is calculated using the distance formula:
  $Distance = \sqrt{(x<em>2 - x</em>1)^2 + (y<em>2 - y</em>1)^2}$

• The total approximate arc length is the sum of the lengths of all these line segments.

• The more line segments (larger $n$) used, the better the approximation of the actual arc length will be.

Example 1: Estimate the length of the curve $f(x) = \sqrt{x} + 1$ for $0 \le x \le 3$. Endpoints: $(0,1)$ and $(3,2)$.

• Using five points (four segments, $n=4$): Divide the interval $[0,3]$ into $4$ subintervals of length $\frac{3-0}{4} = 0.75$ at $x = 0, 0.75, 1.5, 2.25, 3$. The corresponding points are $(0,1)$, $(0.75,1.323)$, $(1.5,1.581)$, $(2.25,1.803)$, $(3,2)$.

  • Segment 1: $\sqrt{(1.323-1)^2 + (0.75-0)^2} \approx 0.817$

  • Segment 2: $\sqrt{(1.581-1.323)^2 + (1.5-0.75)^2} \approx 0.793$

  • Segment 3: $\sqrt{(1.803-1.581)^2 + (2.25-1.5)^2} \approx 0.782$

  • Segment 4: $\sqrt{(2-1.803)^2 + (3-2.25)^2} \approx 0.776$

  • Approximate Arc Length ($n=4$) $\approx 0.817 + 0.793 + 0.782 + 0.776 = 3.167$

• Using nine points (eight segments, $n=8$): Divide the interval $[0,3]$ into $8$ subintervals of length $\frac{3-0}{8} = 0.375$. The points are $(0,1)$, $(0.375,1.17)$, $(0.75,1.323)$, $(1.125,1.46)$, $(1.5,1.581)$, $(1.875,1.696)$ (text shows $1.88$, assuming a typo, should be $1.875$ for even spacing), $(2.25,1.803)$, $(2.625,1.904)$ (text shows $2.63$), $(3,2)$.

  • Calculate lengths of 8 segments (as shown in transcript).

  • Approximate Arc Length ($n=8$) $\approx 3.168$

• Actual Arc Length: The actual arc length for this function over this interval is approximately $3.168$. More advanced methods using definite integrals are covered later in calculus.

Exercise Example: Estimate the length of the curve $f(x) = x^2 + 1$ for $-2 \le x \le 2$. Endpoints: $(-2,5)$ and $(2,5)$.

• Using five points (four segments, $n=4$): Divide the interval $[-2,2]$ into $4$ subintervals of length $(2 - (-2))/4 = 1$ at $x = -2, -1, 0, 1, 2$. The corresponding points are $(-2,5)$, $(-1,2)$, $(0,1)$, $(1,2)$, $(2,5)$.

  • Segment 1: $\sqrt{(2-5)^2 + (-1-(-2))^2} = \sqrt{(-3)^2 + (1)^2} = \sqrt{9+1} = \sqrt{10} \approx 3.162$

  • Segment 2: $\sqrt{(1-2)^2 + (0-(-1))^2} = \sqrt{(-1)^2 + (1)^2} = \sqrt{1+1} = \sqrt{2} \approx 1.414$

  • Segment 3: $\sqrt{(2-1)^2 + (1-0)^2} = \sqrt{(1)^2 + (1)^2} = \sqrt{2} \approx 1.414$

  • Segment 4: $\sqrt{(5-2)^2 + (2-1)^2} = \sqrt{(3)^2 + (1)^2} = \sqrt{10} \approx 3.162$

  • Approximate Arc Length ($n=4$) $\approx 3.162 + 1.414 + 1.414 + 3.162 = 9.153$

• Using nine points (eight segments, $n=8$): Divide the interval $[-2,2]$ into $8$ subintervals of length $(2 - (-2))/8 = 0.5$ at $x = -2, -1.5, -1, -0.5, 0, 0.5, 1, 1.5, 2$. Calculate $(x, f(x))$ for each point and sum the distances for 8 segments.

  • (Details provided in transcript, sum of 8 segments)

  • Approximate Arc Length ($n=8$) $\approx 9.253$

• Actual Arc Length: The actual arc length for this function over this interval is approximately $9.2936$.

Example 2: Estimate the arc length of the curve $y = \sin x$ for $0 \le x \le \pi$. Endpoints: $(0,0)$ and $(\pi,0)$.

• Using three points (two segments, $n=2$): $(0,0)$, $(\pi/2, 1)$, $(\pi,0)$.

  • Segment 1: $\sqrt{(1-0)^2 + (\pi/2-0)^2} = \sqrt{1 + (\pi/2)^2}$

  • Segment 2: $\sqrt{(0-1)^2 + (\pi-\pi/2)^2} = \sqrt{(-1)^2 + (\pi/2)^2}$

  • Approximate Arc Length ($n=2$) $\approx \sqrt{1 + (\pi/2)^2} + \sqrt{1 + (\pi/2)^2} = 2\sqrt{1 + (\pi/2)^2} \approx 3.7242$

• Using five points (four segments, $n=4$): $(0,0)$, $(\pi/4, 1/\sqrt{2})$, $(\pi/2, 1)$, $(3\pi/4, 1/\sqrt{2})$, $(\pi,0)$.

  • The calculation involves summing the lengths of these four segments. Due to symmetry, $\text{d1} = d4$ and $\text{d2} = d3$. The formula is given as:
    $d_4 = \sqrt{(\frac{1}{\sqrt{2}}-0)^2 + (\frac{\pi}{4}-0)^2} + \sqrt{(1-\frac{1}{\sqrt{2}})^2 + (\frac{\pi}{2}-\frac{\pi}{4})^2} + \sqrt{(\frac{1}{\sqrt{2}}-1)^2 + (\frac{3\pi}{4}-\frac{\pi}{2})^2} + \sqrt{(0-\frac{1}{\sqrt{2}})^2 + (\pi-\frac{3\pi}{4})^2} \approx 3.7901$

  • Approximate Arc Length ($n=4$) $\approx 3.7901$

  • A simplified calculation is shown using symmetry: $2\sqrt{(\frac{1}{\sqrt{2}}-0)^2 + (\frac{\pi}{4}-0)^2} + 2\sqrt{(\frac{\pi}{2}-\frac{\pi}{4})^2 + (1-\frac{1}{\sqrt{2}})^2} \approx 3.7901$

Exercise Example: Estimate the arc length of the curve $f(x) = \cos x$ for $0 \le x \le \frac{\pi}{2}$. Endpoints: $(0,1)$ and $(\frac{\pi}{2},0)$.

• Using five points (four segments, $n=4$): Divide $[0,\pi/2]$ into $4$ subintervals of length $\frac{\pi/2 - 0}{4} = \frac{\pi}{8}$. Points: $(0,1)$, $(\pi/8, 0.92)$, $(\pi/4, 0.71)$, $(3\pi/8, 0.383)$, $(\pi/2,0)$.

  • Segment 1: $\sqrt{(0.92-1)^2 + (\pi/8-0)^2} \approx 0.4$

  • Segment 2: $\sqrt{(0.71-0.92)^2 + (\pi/4-\pi/8)^2} \approx 0.449$

  • Segment 3: $\sqrt{(0.383-0.71)^2 + (3\pi/8-\pi/4)^2} \approx 0.509$

  • Segment 4: $\sqrt{(0-0.383)^2 + (\pi/2-3\pi/8)^2} \approx 0.548$

  • Approximate Arc Length ($n=4$) $\approx 0.4 + 0.449 + 0.509 + 0.548 = 1.906$

• Using nine points (eight segments, $n=8$): Divide $[0,\pi/2]$ into $8$ subintervals of length $\frac{\pi/2 - 0}{8} = \frac{\pi}{16}$. Points given with their $y$ values. (Details provided in transcript).

  • Approximate Arc Length ($n=8$) $\approx 1.909$

• Actual Arc Length: The actual arc length is approximately $1.909$.

Exercise Example: Estimate the length of the curve $f(x) = x^2 - x$ for $0 \le x \le 2$. Endpoints: $(0,0)$ and $(2,2)$.

• Using five points (four segments, $n=4$): Divide $[0,2]$ into $4$ subintervals of length $\frac{2-0}{4} = 0.5$. Points: $(0,0)$, $(0.5,-0.25)$, $(1,0)$, $(1.5,0.75)$, $(2,2)$.

  • Segment 1: $\sqrt{(0.5-0)^2 + (-0.25-0)^2} \approx 0.5590$

  • Segment 2: $\sqrt{(1-0.5)^2 + (0-(-0.25))^2} \approx 0.5590$

  • Segment 3: $\sqrt{(1.5-1)^2 + (0.75-0)^2} \approx 0.9014$

  • Segment 4: $\sqrt{(2-1.5)^2 + (2-0.75)^2} \approx 1.3463$

  • Approximate Arc Length ($n=4$) $\approx 0.5590 + 0.5590 + 0.9014 + 1.3463 = 3.3657$

Exercise Example: Estimate the length of the curve $y = \sqrt{1-x^2}$ for $0 \le x \le 1$. Endpoints: $(0,1)$ and $(1,0)$.

• Using five points (four segments, $n=4$): Divide $[0,1]$ into $4$ subintervals of length $\frac{1-0}{4} = 0.25$. Points: $(0,1)$, $(0.25,0.9682)$, $(0.5,0.866)$, $(0.75,0.661)$, $(1,0)$.

  • Segment 1: $\sqrt{(0.9682-1)^2 + (0.25-0)^2} \approx 0.2520$

  • Segment 2: $\sqrt{(0.866-0.9682)^2 + (0.5-0.25)^2} \approx 0.2773$

  • Segment 3: $\sqrt{(0.661-0.866)^2 + (0.75-0.5)^2} \approx 0.3200$

  • Segment 4: $\sqrt{(0-0.661)^2 + (1-0.75)^2} \approx 0.7066$

  • Approximate Arc Length ($n=4$) $\approx 0.2520 + 0.2773 + 0.3200 + 0.7066 = 1.5559$

• Observation: The function $y = \sqrt{1-x^2}$ represents a circle of radius $1$ centered at the origin ($x^2 + y^2 = 1$). The interval $0 \le x \le 1$ and $y \ge 0$ describes the portion of the circle in the first quadrant, which is one-quarter of the full circle's circumference.

  • The circumference of a circle is $2\pi r$.

  • For this curve, $r = 1$.

  • The actual arc length is $\frac{1}{4} (2\pi r) = \frac{2\pi(1)}{4} = \frac{\pi}{2} \approx 1.5708$.

Plenary Questions and Homework

• Review of learning objectives.

• Homework assignments from Alef Lesson 16 and Lesson 18, and exercises from pages 67-68 of the textbook (Q2, Q4, Q6, Q8, Q10, Q12).

Plenary Example: Estimate the arc length of the curve $f(x) = x^3 + 2$ on the interval $-1 \le x \le 1$ using $n=2$ line segments.

• The interval length is $1 - (-1) = 2$. With $n=2$ segments, each segment has an $x$-width of $\frac{2}{2} = 1$. The points are at $x = -1, 0, 1$.

• Points on the curve:

  • When $x = -1$: $f(-1) = (-1)^3 + 2 = -1 + 2 = 1$. Point (-1,1).

  • When $x = 0$: $f(0) = (0)^3 + 2 = 2$. Point (0,2).

  • When $x = 1$: $f(1) = (1)^3 + 2 = 1 + 2 = 3$. Point (1,3).

• Segment 1 (from $(-1,1)$ to $(0,2)$) length: $\sqrt{(0 - (-1))^2 + (2-1)^2} = \sqrt{(1)^2 + (1)^2} = \sqrt{1+1} = \sqrt{2}$

• Segment 2 (from $(0,2)$ to $(1,3)$) length: $\sqrt{(1-0)^2 + (3-2)^2} = \sqrt{(1)^2 + (1)^2} = \sqrt{1+1} = \sqrt{2}$

• Total Arc Length $= \sqrt{2} + \sqrt{2} = 2\sqrt{2}$.

  • The correct answer is Option A ($2\sqrt{2}$).

Homework Answers (Pages 67-68)

Slope Estimation:

• Q2: $f(x) = x^3 + 2$

  • (a) $a = 1$: Answer: $3$

  • (b) $a = 2$: Answer: $12$

• Q4: $f(x) = \sqrt{x+1}$

  • (a) $a = 0$: Answer: $0.5$

  • (b) $a = 3$: Answer: $0.25$

• Q6: $f(x) = \ln x$

  • (a) $a = 1$: Answer: $1$

  • (b) $a = 2$: Answer: $1/2$

Arc Length Estimation:

• Q8: $f(x) = \sin x$, $0 \le x \le \pi/2$

  • (a) $n=4$: Answer: $1.906$

  • (b) $n=8$: Answer: $1.909$

• Q10: $f(x) = 1/x$, $1 \le x \le 2$

  • (a) $n=4$: Answer: $1.131$

  • (b) $n=8$: Answer: $1.132$

• Q12: $f(x) = x^3 + 2$, $-1 \le x \le 1$

  • (a) $n=4$: Answer: $3.0463$

  • (b) $n=8$: Answer: $3.084$