Chap 5. Application of differentiation

# Chap 5. Application of differentiation ## Objectifs et prérequis This chapter applies differentiation to analyze the behavior of functions through gradients, tangents, normals, stationary points, curve sketching, and rates of change. Students will learn to interpret derivatives geometrically and physically, solve optimization problems, and model real-world scenarios involving change. **Prerequisites**: - Proficiency in differentiation rules (power, product, quotient, chain) for algebraic, trigonometric, exponential, and logarithmic functions. - Ability to compute first and second derivatives. - Familiarity with coordinate geometry (slope, line equations). - Understanding of function domains, ranges, and continuity. --- ## I. Gradient, Tangent and Normal ### Definition and Geometric Interpretation The **gradient** of a function $y = f(x)$ at a point $x = a$ is the value of its derivative $f'(a)$, representing the instantaneous rate of change and the slope of the tangent line to the curve at that point. > **To retain**: The gradient of $y = f(x)$ at $x = a$ is $m = f'(a)$. The **tangent line** to the curve at $(a, f(a))$ is the straight line that touches the curve at that point and has the same gradient as the curve there. Its equation is: $$y - f(a) = f'(a)(x - a)$$ The **normal line** is perpendicular to the tangent at the point of contact. Since the product of gradients of perpendicular lines is $-1$, if the tangent has gradient $m$, the normal has gradient $-\frac{1}{m}$ (provided $m \neq 0$). Its equation is: $$y - f(a) = -\frac{1}{f'(a)}(x - a)$$ ### Worked Example 1 Find the equations of the tangent and normal to the curve $y = x^3 - 2x + 1$ at $x = 2$. **Step 1**: Compute $f(2)$: $f(2) = (2)^3 - 2(2) + 1 = 8 - 4 + 1 = 5$ → Point: $(2, 5)$ **Step 2**: Differentiate: $f'(x) = 3x^2 - 2$ **Step 3**: Evaluate gradient at $x = 2$: $f'(2) = 3(4) - 2 = 12 - 2 = 10$ → Tangent gradient $m_T = 10$ **Step 4**: Tangent equation: $y - 5 = 10(x - 2)$ → $y = 10x - 15$ **Step 5**: Normal gradient: $m_N = -\frac{1}{10}$ Normal equation: $y - 5 = -\frac{1}{10}(x - 2)$ → Multiply by 10: $10y - 50 = -x + 2$ → $x + 10y = 52$ ### Worked Example 2 Find the point on the curve $y = x^2 - 4x + 5$ where the tangent is parallel to the line $2x - y + 3 = 0$. **Step 1**: Rewrite line in slope-intercept form: $y = 2x + 3$ → Gradient $m = 2$ **Step 2**: Differentiate curve: $f'(x) = 2x - 4$ **Step 3**: Set derivative equal to 2: $2x - 4 = 2$ → $2x = 6$ → $x = 3$ **Step 4**: Find $y$: $y = (3)^2 - 4(3) + 5 = 9 - 12 + 5 = 2$ → Point: $(3, 2)$ ### Exercises (Corrected) 1. Find the tangent and normal to $y = \frac{1}{x}$ at $x = -2$. - $f(-2) = -\frac{1}{2}$ - $f'(x) = -\frac{1}{x^2}$ → $f'(-2) = -\frac{1}{4}$ - Tangent: $y + \frac{1}{2} = -\frac{1}{4}(x + 2)$ → $y = -\frac{1}{4}x - 1$ - Normal gradient: $4$ → Normal: $y + \frac{1}{2} = 4(x + 2)$ → $y = 4x + \frac{15}{2}$ 2. Where is the tangent to $y = \sin x$ horizontal on $[0, 2\pi]$? - $f'(x) = \cos x = 0$ → $x = \frac{\pi}{2}, \frac{3\pi}{2}$ - Points: $\left(\frac{\pi}{2}, 1\right)$, $\left(\frac{3\pi}{2}, -1\right)$ --- ## II. Stationary Points ### Definition and Classification A **stationary point** occurs where the gradient of the function is zero: $f'(x) = 0$. These points may be local maxima, local minima, or points of inflection. > **To retain**: Stationary points satisfy $f'(x) = 0$. To classify a stationary point at $x = a$: - Compute $f''(a)$: - If $f''(a) > 0$ → local minimum - If $f''(a) < 0$ → local maximum - If $f''(a) = 0$ → test further (sign change of $f'(x)$ around $a$) ### Worked Example 1 Find and classify the stationary points of $y = x^3 - 3x^2 - 9x + 5$. **Step 1**: Differentiate: $f'(x) = 3x^2 - 6x - 9$ **Step 2**: Set $f'(x) = 0$: $3x^2 - 6x - 9 = 0$ → Divide by 3: $x^2 - 2x - 3 = 0$ → $(x-3)(x+1)=0$ → $x = 3$ or $x = -1$ **Step 3**: Second derivative: $f''(x) = 6x - 6$ **Step 4**: Evaluate: - At $x = -1$: $f''(-1) = -6 - 6 = -12 < 0$ → local maximum $y = (-1)^3 - 3(-1)^2 - 9(-1) + 5 = -1 - 3 + 9 + 5 = 10$ → $(-1, 10)$ - At $x = 3$: $f''(3) = 18 - 6 = 12 > 0$ → local minimum $y = 27 - 27 - 27 + 5 = -22$ → $(3, -22)$ ### Worked Example 2 Analyze $y = x^4$ at $x = 0$. **Step 1**: $f'(x) = 4x^3$ → $f'(0) = 0$ → stationary point **Step 2**: $f''(x) = 12x^2$ → $f''(0) = 0$ → inconclusive **Step 3**: Check sign of $f'(x)$: - For $x < 0$ (e.g., $x = -0.1$): $f'(-0.1) = 4(-0.001) = -0.004 < 0$ - For $x > 0$ (e.g., $x = 0.1$): $f'(0.1) = 0.004 > 0$ → Derivative changes from negative to positive → local minimum at $(0, 0)$ ### Exercises (Corrected) 1. Find and classify stationary points of $y = 2x^3 - 9x^2 + 12x - 4$. - $f'(x) = 6x^2 - 18x + 12 = 6(x^2 - 3x + 2) = 6(x-1)(x-2)$ → $x=1,2$ - $f''(x) = 12x - 18$ - $x=1$: $f''(1) = -6 < 0$ → max; $y=2-9+12-4=1$ → $(1,1)$ - $x=2$: $f''(2)=6>0$ → min; $y=16-36+24-4=0$ → $(2,0)$ 2. Show that $y = x + \frac{1}{x}$ ($x>0$) has a minimum at $x=1$. - $f'(x) = 1 - \frac{1}{x^2}$ → Set to 0: $x^2=1$ → $x=1$ (since $x>0$) - $f''(x) = \frac{2}{x^3}$ → $f''(1)=2>0$ → minimum - $y=1+1=2$ → Minimum value 2 at $(1,2)$ --- ## III. Elementary Curve Sketching ### Strategy for Sketching $y = f(x)$ To sketch a curve effectively: 1. **Domain**: Identify where $f(x)$ is defined. 2. **Intercepts**: - $y$-intercept: $f(0)$ - $x$-intercepts: solve $f(x)=0$ 3. **Symmetry**: - Even: $f(-x)=f(x)$ → symmetric about $y$-axis - Odd: $f(-x)=-f(x)$ → symmetric about origin 4. **Asymptotes**: - Vertical: where $f(x) \to \pm\infty$ (e.g., denominator zero in rational functions) - Horizontal: $\lim_{x\to\pm\infty} f(x) = L$ - Oblique: if degree(num) = degree(den)+1 in rational functions 5. **Stationary points**: Solve $f'(x)=0$, classify via $f''(x)$ or sign test. 6. **Behavior**: Use first and second derivatives to determine increasing/decreasing and concavity. 7. **Plot key points and sketch**. > **To retain**: Curve sketching uses intercepts, symmetry, asymptotes, stationary points, and derivative signs. ### Worked Example 1 Sketch $y = \frac{x^2}{x^2 - 4}$. **Step 1**: Domain: $x^2 - 4 \neq 0$ → $x \neq \pm 2$ **Step 2**: Intercepts: - $y$-int: $f(0) = 0$ → $(0,0)$ - $x$-int: $x^2=0$ → $x=0$ → same point **Step 3**: Symmetry: $f(-x) = \frac{(-x)^2}{(-x)^2-4} = \frac{x^2}{x^2-4} = f(x)$ → even → symmetric about $y$-axis **Step 4**: Asymptotes: - Vertical: $x = \pm 2$ (denominator zero, numerator non-zero) - Horizontal: $\lim_{x\to\pm\infty} \frac{x^2}{x^2-4} = 1$ → $y=1$ **Step 5**: Derivatives: $f'(x) = \frac{(2x)(x^2-4) - x^2(2x)}{(x^2-4)^2} = \frac{2x^3 - 8x - 2x^3}{(x^2-4)^2} = \frac{-8x}{(x^2-4)^2}$ Set $f'(x)=0$ → $-8x=0$ → $x=0$ $f''(x)$: complex, but sign of $f'(x)$: - $x < -2$: $f'(x) = \frac{-8(-)}{(+)} = +$ → increasing - $-2 < x < 0$: $f'(x) = \frac{-8(-)}{(+)} = +$ → increasing - $0 < x < 2$: $f'(x) = \frac{-8(+)}{(+)} = -$ → decreasing - $x > 2$: $f'(x) = \frac{-8(+)}{(+)} = -$ → decreasing → $x=0$ is local maximum? Wait: increasing then decreasing → yes, local max at $(0,0)$ But $f(0)=0$, and $y\to 1$ as $x\to\pm\infty$, so actually $(0,0)$ is a local max? Check values: $f(1)=1/(1-4)=-1/3 <0$, so yes, $(0,0)$ is local max. **Step 6**: Behavior near asymptotes: - As $x\to 2^-$: denominator $0^-$, numerator $+$ → $y\to -\infty$ - As $x\to 2^+$: denominator $0^+$, numerator $+$ → $y\to +\infty$ - Symmetric for $x\to -2$ **Step 7**: Sketch: - Symmetric about $y$-axis - Vertical asymptotes at $x=\pm 2$ - Horizontal asymptote $y=1$ - Passes through $(0,0)$, local max - For $x>2$: decreasing from $+\infty$ to $1$ - For $0<x<2$: decreasing from $0$ to $-\infty$ - etc. ### Worked Example 2 Sketch $y = x^3 - 3x$. **Step 1**: Domain: all real numbers **Step 2**: Intercepts: - $y$-int: $0$ - $x$-int: $x(x^2-3)=0$ → $x=0, \pm\sqrt{3}$ **Step 3**: Symmetry: $f(-x) = -x^3 + 3x = -(x^3-3x) = -f(x)$ → odd → symmetric about origin **Step 4**: Asymptotes: none (polynomial) **Step 5**: Derivatives: $f'(x) = 3x^2 - 3 = 3(x^2-1)$ → zeros at $x=\pm 1$ $f''(x) = 6x$ **Step 6**: Classify stationary points: - $x=-1$: $f''(-1)=-6<0$ → local max; $y=(-1)^3-3(-1)=-1+3=2$ → $(-1,2)$ - $x=1$: $f''(1)=6>0$ → local min; $y=1-3=-2$ → $(1,-2)$ **Step 7**: Sign of $f'(x)$: - $x < -1$: $f'(x) > 0$ → increasing - $-1 < x < 1$: $f'(x) < 0$ → decreasing - $x > 1$: $f'(x) > 0$ → increasing → Confirms max at $x=-1$, min at $x=1$ **Step 8**: Concavity: $f''(x)=6x$ - $x<0$: $f''<0$ → concave down - $x>0$: $f''>0$ → concave up → Inflection point at $x=0$ ($f''(0)=0$ and sign change) **Step 9**: Sketch: - Passes through $(0,0)$, $(-\sqrt{3},0)$, $(\sqrt{3},0)$ - Local max $(-1,2)$, local min $(1,-2)$ - Inflection at $(0,0)$ - As $x\to -\infty$, $y\to -\infty$; as $x\to +\infty$, $y\to +\infty$ - Symmetric about origin ### Exercises (Corrected) 1. Sketch $y = \frac{1}{x^2 + 1}$. - Domain: all reals - Intercepts: $y$-int $(0,1)$; no $x$-int (numerator 1≠0) - Symmetry: even - Asymptotes: horizontal $y=0$ (as $x\to\pm\infty$) - $f'(x) = -\frac{2x}{(x^2+1)^2}$ → zero at $x=0$ - $x<0$: $f'(x)>0$ → increasing; $x>0$: $f'(x)<0$ → decreasing → max at $(0,1)$ - $f''(x)$: positive for $|x|>1/\sqrt{3}$, negative for $|x|<1/\sqrt{3}$ → inflection at $x=\pm 1/\sqrt{3}$ - Bell-shaped curve, max at $(0,1)$, approaches 0 at ends. 2. Sketch $y = x^4 - 4x^3$. - Factor: $x^3(x-4)$ - Intercepts: $x$-int: $0,4$; $y$-int: $0$ - Symmetry: none - $f'(x)=4x^3-12x^2=4x^2(x-3)$ → zeros at $x=0$ (double), $x=3$ - $f''(x)=12x^2-24x=12x(x-2)$ - $x=0$: $f''(0)=0$ → test $f'$: $x<0$: $f'(x)=4(+)^2(-)=-$; $0<x<3$: $f'(x)=4(+)^2(-)=-$ → no change → not extremum (terrace point) - $x=3$: $f''(3)=12(3)(1)>0$ → min; $y=81-108=-27$ → $(3,-27)$ - As $x\to\pm\infty$, $y\to+\infty$ - Sketch: crosses at $(0,0)$ and $(4,0)$, terrace at $(0,0)$, min at $(3,-27)$ --- ## IV. Rates of Change ### Definition and Applications The derivative $\frac{dy}{dx}$ represents the **instantaneous rate of change** of $y$ with respect to $x$. In related rates problems, we differentiate an equation linking variables with respect to time $t$ to find how one rate affects another. > **To retain**: If $y = f(x)$, then $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$ (chain rule for related rates). ### Worked Example 1 The radius $r$ of a circular oil spill increases at 2 cm/s. How fast is the area $A$ increasing when $r = 5$ cm? **Step 1**: Area of circle: $A = \pi r^2$ **Step 2**: Differentiate both sides with respect to $t$: $\frac{dA}{dt} = \pi \cdot 2r \cdot \frac{dr}{dt} = 2\pi r \frac{dr}{dt}$ **Step 3**: Substitute known values: $\frac{dr}{dt} = 2$ cm/s, $r = 5$ cm $\frac{dA}{dt} = 2\pi (5)(2) = 20\pi$ cm²/s ### Worked Example 2 A ladder 5 m long leans against a vertical wall. The bottom slides away from the wall at 0.5 m/s. How fast is the top descending when the bottom is 3 m from the wall? **Step 1**: Let $x$ = distance from wall to bottom, $y$ = height of top on wall. By Pythagoras: $x^2 + y^2 = 5^2 = 25$ **Step 2**: Differentiate implicitly with respect to $t$: $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$ → $x \frac{dx}{dt} + y \frac{dy}{dt} = 0$ **Step 3**: Known: $\frac{dx}{dt} = 0.5$ m/s, $x = 3$ m Find $y$: $3^2 + y^2 = 25$ → $y^2 = 16$ → $y = 4$ m (positive since height) **Step 4**: Substitute: $(3)(0.5) + (4) \frac{dy}{dt} = 0$ → $1.5 + 4 \frac{dy}{dt} = 0$ → $\frac{dy}{dt} = -\frac{1.5}{4} = -0.375$ m/s Negative sign indicates $y$ is decreasing → top descending at 0.375 m/s ### Worked Example 3 Water is poured into an inverted conical tank at 10 m³/min. The tank has height 6 m and base radius 3 m. How fast is the water level rising when the water is 4 m deep? **Step 1**: Volume of cone: $V = \frac{1}{3} \pi r^2 h$ By similar triangles: $\frac{r}{h} = \frac{3}{6} = \frac{1}{2}$ → $r = \frac{h}{2}$ **Step 2**: Substitute: $V = \frac{1}{3} \pi \left(\frac{h}{2}\right)^2 h = \frac{1}{3} \pi \cdot \frac{h^2}{4} \cdot h = \frac{\pi}{12} h^3$ **Step 3**: Differentiate w.r.t. $t$: $\frac{dV}{dt} = \frac{\pi}{12} \cdot 3h^2 \frac{dh}{dt} = \frac{\pi}{4} h^2 \frac{dh}{dt}$ **Step 4**: Known: $\frac{dV}{dt} = 10$ m³/min, $h = 4$ m $10 = \frac{\pi}{4} (4)^2 \frac{dh}{dt} = \frac{\pi}{4} \cdot 16 \cdot \frac{dh}{dt} = 4\pi \frac{dh}{dt}$ → $\frac{dh}{dt} = \frac{10}{4\pi} = \frac{5}{2\pi}$ m/min ≈ 0.796 m/min ### Exercises (Corrected) 1. The side of a cube increases at 0.1 cm/s. Find the rate of increase of its volume when the side is 5 cm. - $V = s^3$ → $\frac{dV}{dt} = 3s^2 \frac{ds}{dt}$ - $\frac{ds}{dt}=0.1$, $s=5$ → $\frac{dV}{dt}=3(25)(0.1)=7.5$ cm³/s 2. A spherical balloon is inflated at 100 cm³/s. How fast is the radius increasing when the radius is 10 cm? - $V = \frac{4}{3}\pi r^3$ → $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$ - $100 = 4\pi (100) \frac{dr}{dt}$ → $\frac{dr}{dt} = \frac{100}{400\pi} = \frac{1}{4\pi}$ cm/s 3. A particle moves along $y = x^2 - 4x + 5$. If $\frac{dx}{dt} = 3$ units/s, find $\frac{dy}{dt}$ when $x = 2$. - $\frac{dy}{dt} = (2x - 4) \frac{dx}{dt}$ - At $x=2$: $2(2)-4=0$ → $\frac{dy}{dt} = 0 \cdot 3 = 0$ --- ## Synthèse du chapitre - **Gradient**: $f'(a)$ gives slope of tangent at $x=a$; tangent: $y-f(a)=f'(a)(x-a)$; normal: $y-f(a)=-\frac{1}{f'(a)}(x-a)$ if $f'(a)\neq 0$. - **Stationary points**: Solve $f'(x)=0$; classify via $f''(x)$: $>0$ min, $<0$ max, $=0$ test sign change of $f'(x)$. - **Curve sketching**: Use domain, intercepts, symmetry, asymptotes, stationary points, and derivative signs for increasing/decreasing and concavity. - **Rates of change**: $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$; apply chain rule to geometric/physical formulas (area, volume, Pythagoras, etc.). --- ## Glossaire - **Gradient**: The derivative $f'(x)$, representing the slope of the tangent line to $y=f(x)$ at a point. - **Tangent line**: The line that touches a curve at a point and has the same gradient as the curve there. - **Normal line**: The line perpendicular to the tangent at the point of contact. - **Stationary point**: A point where $f'(x)=0$; may be local max, min, or inflection. - **Local maximum**: A point where $f(x)$ is greater than nearby values; $f''(x)<0$ if test applicable. - **Local minimum**: A point where $f(x)$ is less than nearby values; $f''(x)>0$ if test applicable. - **Point of inflection**: Where concavity changes; $f''(x)=0$ and sign change of $f''(x)$ (or $f'(x)$ changes from increasing to decreasing). - **Related rates**: Problems where rates of change of linked variables are found via differentiation with respect to time. - **Critical point**: A point where $f'(x)=0$ or $f'(x)$ is undefined (includes stationary points and points where derivative does not exist). - **Concavity**: Describes whether a curve bends upward ($f''>0$) or downward ($f''<0$). - **Asymptote**: A line that the curve approaches as $x$ or $y$ tends to infinity (vertical, horizontal, or oblique). - **Extremum**: A local maximum or minimum value of a function.