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Calculus - Applications of Differentiation (Tangents, Normals, Optimization, Kinematics)

Grade 12IB

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

🔑Concepts

The Gradient as a Derivative: The derivative dydx\frac{dy}{dx} represents the instantaneous rate of change of a function. Geometrically, this is the gradient of the tangent to the curve at any given point (x,y)(x, y). If the graph is viewed as a path, the tangent line indicates the direction of travel at that specific moment.

Tangents and Normals: A tangent is a straight line that just touches a curve at a point, sharing the same gradient as the curve at that point. A normal is a line perpendicular to the tangent at the point of contact. On a graph, the normal and tangent intersect at a 9090^{\circ} angle, and their gradients are negative reciprocals of each other.

Increasing and Decreasing Functions: A function is strictly increasing on an interval if f(x)>0f'(x) > 0, shown visually as the graph rising from left to right. Conversely, a function is strictly decreasing if f(x)<0f'(x) < 0, shown as the graph falling from left to right. Stationary points occur where f(x)=0f'(x) = 0, where the graph momentarily 'flattens out'.

Local Extrema and Stationary Points: Local maximums and minimums occur where the derivative is zero or undefined. A local maximum appears as a 'peak' on a graph where the gradient changes from positive to negative. A local minimum appears as a 'valley' where the gradient changes from negative to positive. The Second Derivative Test can confirm this: f(x)<0f''(x) < 0 indicates a maximum (concave down), while f(x)>0f''(x) > 0 indicates a minimum (concave up).

Concavity and Points of Inflection: Concavity describes the 'bend' of a curve. A curve is concave up (like a cup) if f(x)>0f''(x) > 0 and concave down (like a cap) if f(x)<0f''(x) < 0. A point of inflection is a point where the concavity changes (e.g., from cup-shaped to cap-shaped). At these points, f(x)=0f''(x) = 0 and there is a sign change in the second derivative.

Optimization: This involves finding the absolute maximum or minimum value of a function within a specific context, such as maximizing the area of a shape or minimizing the cost of production. Visually, this requires identifying the highest or lowest point on the function's graph within the feasible domain, often by checking stationary points and interval endpoints.

Kinematics of a Particle: Differentiation is used to relate displacement s(t)s(t), velocity v(t)v(t), and acceleration a(t)a(t). Velocity is the rate of change of displacement v(t)=s(t)v(t) = s'(t), and acceleration is the rate of change of velocity a(t)=v(t)=s(t)a(t) = v'(t) = s''(t). On a displacement-time graph, the gradient represents velocity; on a velocity-time graph, the gradient represents acceleration.

📐Formulae

Equation of a Tangent: yy1=f(x1)(xx1)y - y_1 = f'(x_1)(x - x_1)

Equation of a Normal: yy1=1f(x1)(xx1)y - y_1 = -\frac{1}{f'(x_1)}(x - x_1)

Gradient Relationship: mtangent×mnormal=1m_{tangent} \times m_{normal} = -1

Velocity: v(t)=dsdtv(t) = \frac{ds}{dt}

Acceleration: a(t)=dvdt=d2sdt2a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2}

Speed: Speed=v(t)\text{Speed} = |v(t)|

Second Derivative Test: If f(c)=0 and f(c)<0, then x=c is a local maximum.\text{If } f'(c) = 0 \text{ and } f''(c) < 0, \text{ then } x=c \text{ is a local maximum.}

Second Derivative Test: If f(c)=0 and f(c)>0, then x=c is a local minimum.\text{If } f'(c) = 0 \text{ and } f''(c) > 0, \text{ then } x=c \text{ is a local minimum.}

💡Examples

Problem 1:

Find the equation of the normal to the curve f(x)=x33x+2f(x) = x^3 - 3x + 2 at the point where x=2x = 2.

Solution:

  1. Find the yy-coordinate: f(2)=233(2)+2=86+2=4f(2) = 2^3 - 3(2) + 2 = 8 - 6 + 2 = 4. So, the point is (2,4)(2, 4).
  2. Find the derivative to get the gradient of the tangent: f(x)=3x23f'(x) = 3x^2 - 3.
  3. Calculate the gradient of the tangent at x=2x = 2: mt=f(2)=3(2)23=123=9m_t = f'(2) = 3(2)^2 - 3 = 12 - 3 = 9.
  4. Find the gradient of the normal: mn=1mt=19m_n = -\frac{1}{m_t} = -\frac{1}{9}.
  5. Use the point-slope formula: y4=19(x2)y - 4 = -\frac{1}{9}(x - 2).
  6. Simplify to general form: 9y36=x+2x+9y38=09y - 36 = -x + 2 \Rightarrow x + 9y - 38 = 0.

Explanation:

To find the normal, we first determine the gradient of the tangent by differentiating the function. Since the normal is perpendicular to the tangent, we take the negative reciprocal of the tangent's gradient and then use the point-slope equation of a line.

Problem 2:

A particle moves along a straight line such that its displacement ss (in meters) at time tt (in seconds) is given by s(t)=2t39t2+12t+5s(t) = 2t^3 - 9t^2 + 12t + 5 for t0t \ge 0. Find the velocity of the particle when its acceleration is zero.

Solution:

  1. Find the velocity function by differentiating displacement: v(t)=s(t)=6t218t+12v(t) = s'(t) = 6t^2 - 18t + 12.
  2. Find the acceleration function by differentiating velocity: a(t)=v(t)=12t18a(t) = v'(t) = 12t - 18.
  3. Set acceleration to zero to find the time: 12t18=012t=18t=1.512t - 18 = 0 \Rightarrow 12t = 18 \Rightarrow t = 1.5 seconds.
  4. Calculate velocity at t=1.5t = 1.5: v(1.5)=6(1.5)218(1.5)+12v(1.5) = 6(1.5)^2 - 18(1.5) + 12.
  5. v(1.5)=6(2.25)27+12=13.527+12=1.5v(1.5) = 6(2.25) - 27 + 12 = 13.5 - 27 + 12 = -1.5 ms1m s^{-1}.

Explanation:

Kinematics problems require successive differentiation. We differentiate displacement to find velocity, and velocity to find acceleration. We then solve for the specific time tt when acceleration is zero and substitute that back into the velocity expression.