krit.club logo

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

•

Rate of Change of Quantities: The derivative dydx\frac{dy}{dx} represents the instantaneous rate of change of yy with respect to xx. Visually, if we plot a graph of a quantity like displacement against time, the slope of the curve at any specific point represents the velocity at that instant. If dydx>0\frac{dy}{dx} > 0, the quantity is increasing; if dydx<0\frac{dy}{dx} < 0, it is decreasing.

•

Tangents to a Curve: A tangent at a point P(x1,y1)P(x_1, y_1) is a straight line that 'skims' the curve, touching it exactly at that point. Geometrically, the slope of this tangent line is equal to the value of the derivative f′(x)f'(x) evaluated at x1x_1. If the tangent is horizontal (parallel to the x-axis), the slope is zero.

•

Normals to a Curve: The normal is a straight line perpendicular to the tangent at the point of contact P(x1,y1)P(x_1, y_1). Visually, it stands at a 90∘90^{\circ} angle to the curve's surface at that point. Since the product of the slopes of two perpendicular lines is −1-1, the slope of the normal is the negative reciprocal of the derivative at that point.

•

Increasing and Decreasing Functions: A function is strictly increasing on an interval if its graph continuously rises from left to right, which corresponds to f′(x)>0f'(x) > 0 for all xx in that interval. Conversely, it is strictly decreasing if the graph falls from left to right, which corresponds to f′(x)<0f'(x) < 0. If f′(x)=0f'(x) = 0 throughout an interval, the function is constant (a horizontal line).

•

Local Maxima and Minima (Turning Points): These are points where the function reaches a peak or a valley. A local maximum occurs at a point where the function's graph changes from rising to falling (the slope changes from positive to negative). A local minimum occurs where the graph changes from falling to rising (the slope changes from negative to positive). At both types of points, the tangent is horizontal, meaning f′(x)=0f'(x) = 0.

•

The Second Derivative Test: This test uses the curvature (concavity) of the graph to identify the nature of critical points. If f′(c)=0f'(c) = 0 and f′′(c)<0f''(c) < 0, the graph is 'concave down' (like an upside-down bowl), indicating a local maximum. If f′(c)=0f'(c) = 0 and f′′(c)>0f''(c) > 0, the graph is 'concave up' (like a valley), indicating a local minimum.

•

Absolute Maxima and Minima: On a closed interval [a,b][a, b], the absolute maximum is the highest point the graph ever reaches, and the absolute minimum is the lowest. These points can occur either at critical points where f′(x)=0f'(x) = 0 or at the boundaries (endpoints) of the interval aa and bb.

📐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).