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Calculus - Applications of Derivatives: Rate of Change, Increasing/Decreasing Functions, Tangents and Normals, Maxima and Minima

Grade 12ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The derivative dydx\frac{dy}{dx} represents the instantaneous rate of change of yy with respect to xx. If ss is displacement and tt is time, the first derivative v=dsdtv = \frac{ds}{dt} is velocity, and the second derivative a=d2sdt2a = \frac{d^2s}{dt^2} is acceleration.

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For a curve y=f(x)y = f(x), the slope of the tangent at a point (x1,y1)(x_1, y_1) is given by m=f′(x1)m = f'(x_1). The normal is perpendicular to the tangent, so its slope is mn=−1f′(x1)m_n = -\frac{1}{f'(x_1)}.

Graph showing a curve with a tangent and normal line at a specific point P.
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A function f(x)f(x) is strictly increasing on an interval if f′(x)>0f'(x) > 0 for all xx in that interval, meaning the graph moves upwards as xx increases. Conversely, it is strictly decreasing if f′(x)<0f'(x) < 0.

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Local maxima and minima (stationary points) occur where f′(x)=0f'(x) = 0. The Second Derivative Test states that if f′(c)=0f'(c) = 0 and f′′(c)<0f''(c) < 0, x=cx=c is a local maximum. If f′′(c)>0f''(c) > 0, x=cx=c is a local minimum.

📐Formulae

Rate of Change: dydx=ddx[f(x)]\frac{dy}{dx} = \frac{d}{dx}[f(x)]

Velocity: v=dsdtv = \frac{ds}{dt} and Acceleration: a=dvdt=d2sdt2a = \frac{dv}{dt} = \frac{d^2s}{dt^2}

Slope of Tangent (mm): m=(dydx)(x1,y1)m = \left(\frac{dy}{dx}\right)_{(x_1, y_1)}

Equation of Tangent: y−y1=m(x−x1)y - y_1 = m(x - x_1)

Slope of Normal (mnm_n): mn=−1m=−1(dydx)(x1,y1)m_n = -\frac{1}{m} = -\frac{1}{\left(\frac{dy}{dx}\right)_{(x_1, y_1)}}

Equation of Normal: y−y1=−1m(x−x1)y - y_1 = -\frac{1}{m}(x - x_1)

Condition for Increasing Function: f′(x)≥0f'(x) \geq 0

Condition for Decreasing Function: f′(x)≤0f'(x) \leq 0

Critical Point Condition: f′(x)=0f'(x) = 0 or f′(x)f'(x) is not defined

💡Examples

Problem 1:

Find the equations of the tangent and the normal to the curve y=x2+4x+1y = x^2 + 4x + 1 at the point where x=1x = 1.

Solution:

Step 1: Find the yy-coordinate at x=1x = 1. y=(1)2+4(1)+1=6y = (1)^2 + 4(1) + 1 = 6. So, the point is (1,6)(1, 6). Step 2: Find the derivative to get the slope. dydx=2x+4\frac{dy}{dx} = 2x + 4. Step 3: Calculate the slope mm at x=1x = 1. m=2(1)+4=6m = 2(1) + 4 = 6. Step 4: Equation of the tangent. Using y−y1=m(x−x1)y - y_1 = m(x - x_1): y−6=6(x−1)  ⟹  y−6=6x−6  ⟹  6x−y=0y - 6 = 6(x - 1) \implies y - 6 = 6x - 6 \implies 6x - y = 0. Step 5: Equation of the normal. Slope of normal mn=−16m_n = -\frac{1}{6}. Using y−y1=mn(x−x1)y - y_1 = m_n(x - x_1): y−6=−16(x−1)  ⟹  6y−36=−x+1  ⟹  x+6y−37=0y - 6 = -\frac{1}{6}(x - 1) \implies 6y - 36 = -x + 1 \implies x + 6y - 37 = 0.

Explanation:

To find tangent and normal equations, first determine the point of contact, then use the derivative to find the tangent's slope. The normal's slope is the negative reciprocal of the tangent's slope.

Problem 2:

Find the local maximum and minimum values of the function f(x)=x3−3x+2f(x) = x^3 - 3x + 2.

Solution:

Step 1: Find the first derivative and set it to zero. f′(x)=3x2−3f'(x) = 3x^2 - 3. Setting f′(x)=0  ⟹  3(x2−1)=0  ⟹  x=1,−1f'(x) = 0 \implies 3(x^2 - 1) = 0 \implies x = 1, -1. Step 2: Use the second derivative test. f′′(x)=6xf''(x) = 6x. Step 3: Test x=1x = 1. f′′(1)=6(1)=6f''(1) = 6(1) = 6. Since f′′(1)>0f''(1) > 0, x=1x = 1 is a point of local minimum. Local minimum value f(1)=(1)3−3(1)+2=0f(1) = (1)^3 - 3(1) + 2 = 0. Step 4: Test x=−1x = -1. f′′(−1)=6(−1)=−6f''(-1) = 6(-1) = -6. Since f′′(−1)<0f''(-1) < 0, x=−1x = -1 is a point of local maximum. Local maximum value f(−1)=(−1)3−3(−1)+2=−1+3+2=4f(-1) = (-1)^3 - 3(-1) + 2 = -1 + 3 + 2 = 4.

Explanation:

Critical points are found where the first derivative is zero. The second derivative test is then applied: a positive result indicates a minimum (valley), and a negative result indicates a maximum (peak).

Problem 3:

A ladder 55 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 22 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 44 m away from the wall?

Right-angled triangle representing a ladder leaning against a wall.

Solution:

Let xx be the distance of the foot from the wall and yy be the height of the ladder on the wall. By Pythagoras theorem: x2+y2=52=25x^2 + y^2 = 5^2 = 25 Differentiating with respect to time tt: 2xdxdt+2ydydt=02x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 Given dxdt=2\frac{dx}{dt} = 2 cm/s =0.02= 0.02 m/s. When x=4x = 4: 42+y2=25  ⟹  y=34^2 + y^2 = 25 \implies y = 3 m. Substitute values: 2(4)(0.02)+2(3)dydt=02(4)(0.02) + 2(3)\frac{dy}{dt} = 0 0.16+6dydt=0  ⟹  dydt=−0.166=−2750.16 + 6\frac{dy}{dt} = 0 \implies \frac{dy}{dt} = -\frac{0.16}{6} = -\frac{2}{75} m/s. The height is decreasing at 83\frac{8}{3} cm/s.

Explanation:

We use the Pythagorean relation to link the horizontal and vertical distances, then use implicit differentiation with respect to time to find the related rate of change.

Problem 4:

Show that the function f(x)=x3−6x2+12x−18f(x) = x^3 - 6x^2 + 12x - 18 is increasing on R\mathbb{R}.

Graph of f(x) = x^3 - 6x^2 + 12x - 18 showing a continuous upward trend.

Solution:

First, find the derivative of the function: f′(x)=3x2−12x+12f'(x) = 3x^2 - 12x + 12 Factor out 33: f′(x)=3(x2−4x+4)f'(x) = 3(x^2 - 4x + 4) This is a perfect square: f′(x)=3(x−2)2f'(x) = 3(x - 2)^2 Since (x−2)2≥0(x - 2)^2 \geq 0 for all x∈Rx \in \mathbb{R}, then f′(x)≥0f'(x) \geq 0 for all real xx. Therefore, f(x)f(x) is an increasing function on R\mathbb{R}.

Explanation:

To prove a function is increasing, we show that its first derivative is non-negative for all values in the domain.

Applications of Derivatives: Rate of Change, Increasing/Decreasing Functions, Tangents and…