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Calculus - Applications of Differentiation (Tangents, Normals, Rates of Change, Optimization)

Grade 11IB

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

🔑Concepts

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The derivative f′(a)f'(a) represents the instantaneous rate of change of a function at x=ax = a. Geometrically, this is the gradient (slope) of the tangent line to the curve at that specific point. Visually, if you zoom in infinitely on a point on a curve, the curve starts to look like this straight tangent line.

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A tangent line touches a curve at a point P(x1,y1)P(x_1, y_1) and has the same gradient as the curve at that point. On a graph, the tangent illustrates the direction the curve is heading at that exact moment, providing a linear approximation of the function near the point of contact.

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The normal to a curve at point PP is the line perpendicular to the tangent at that same point. Visually, if the tangent is horizontal, the normal is vertical. They meet at a 90∘90^{\circ} angle, and the product of their gradients is always −1-1.

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A function is increasing where f′(x)>0f'(x) > 0 and decreasing where f′(x)<0f'(x) < 0. On a coordinate plane, an increasing function slopes upwards from left to right, while a decreasing function slopes downwards.

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Stationary points occur where the gradient is zero, f′(x)=0f'(x) = 0. These points are visual 'turning points' or 'flat spots' on a graph. A local maximum is a peak (the graph stops rising and starts falling), a local minimum is a valley (the graph stops falling and starts rising), and a stationary point of inflection is a flat spot where the graph continues its original direction.

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The nature of a stationary point can be determined using the second derivative test. If f′′(x)>0f''(x) > 0, the graph is concave up (shaped like a cup or UU), indicating a local minimum. If f′′(x)<0f''(x) < 0, the graph is concave down (shaped like a cap or nn), indicating a local maximum.

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Rates of change involve finding how one variable changes with respect to another, often time tt. For example, in kinematics, the derivative of displacement ss with respect to time tt (dsdt\frac{ds}{dt}) gives velocity. Visually, the steeper the displacement-time graph, the higher the velocity.

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Optimization is the process of finding the maximum or minimum value of a function within a given context. This usually involves creating a mathematical model for a real-world scenario, such as maximizing the volume of a box or minimizing the material used for a container, and finding where the derivative of that model equals zero.

📐Formulae

Gradient of tangent: mtan=f′(x1)m_{tan} = f'(x_1)

Equation of Tangent: y−y1=f′(x1)(x−x1)y - y_1 = f'(x_1)(x - x_1)

Gradient of Normal: mnorm=−1f′(x1)m_{norm} = -\frac{1}{f'(x_1)}

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

Stationary Point Condition: f′(x)=0f'(x) = 0

Second Derivative Test for Minima: f′(x)=0f'(x) = 0 and f′′(x)>0f''(x) > 0

Second Derivative Test for Maxima: f′(x)=0f'(x) = 0 and f′′(x)<0f''(x) < 0

Chain Rule for Rates: dydt=dydx×dxdt\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}

💡Examples

Problem 1:

Find the equation of the normal to the curve f(x)=x2−4x+1f(x) = x^2 - 4x + 1 at the point where x=3x = 3.

Solution:

  1. Find the y-coordinate at x=3x = 3: f(3)=(3)2−4(3)+1=9−12+1=−2f(3) = (3)^2 - 4(3) + 1 = 9 - 12 + 1 = -2. So the point is (3,−2)(3, -2).
  2. Find the derivative to get the gradient of the tangent: f′(x)=2x−4f'(x) = 2x - 4.
  3. Calculate the tangent gradient at x=3x = 3: mtan=f′(3)=2(3)−4=2m_{tan} = f'(3) = 2(3) - 4 = 2.
  4. Find the gradient of the normal: mnorm=−1mtan=−12m_{norm} = -\frac{1}{m_{tan}} = -\frac{1}{2}.
  5. Use the point-slope formula for the normal: y−(−2)=−12(x−3)⇒y+2=−12x+32y - (-2) = -\frac{1}{2}(x - 3) \Rightarrow y + 2 = -\frac{1}{2}x + \frac{3}{2}.
  6. Simplify to standard form: y=−12x−12y = -\frac{1}{2}x - \frac{1}{2} or x+2y+1=0x + 2y + 1 = 0.

Explanation:

This approach first identifies the specific point on the curve, then uses the derivative to find the slope of the curve. Since the normal is perpendicular to the tangent, we use the negative reciprocal of the tangent's slope to find the normal's equation.

Problem 2:

A closed rectangular box has a square base of side length xx cm and a total surface area of 600600 cm2cm^2. Find the value of xx that maximizes the volume VV of the box.

Solution:

  1. Express Surface Area SS in terms of xx and height hh: S=2x2+4xh=600S = 2x^2 + 4xh = 600.
  2. Solve for hh: 4xh=600−2x2⇒h=600−2x24x=150x−x24xh = 600 - 2x^2 \Rightarrow h = \frac{600 - 2x^2}{4x} = \frac{150}{x} - \frac{x}{2}.
  3. Write the Volume formula: V=x2h=x2(150x−x2)=150x−12x3V = x^2h = x^2(\frac{150}{x} - \frac{x}{2}) = 150x - \frac{1}{2}x^3.
  4. Differentiate VV with respect to xx: dVdx=150−32x2\frac{dV}{dx} = 150 - \frac{3}{2}x^2.
  5. Set the derivative to zero for optimization: 150−32x2=0⇒32x2=150⇒x2=100⇒x=10150 - \frac{3}{2}x^2 = 0 \Rightarrow \frac{3}{2}x^2 = 150 \Rightarrow x^2 = 100 \Rightarrow x = 10 (since x>0x > 0).
  6. Verify it is a maximum using the second derivative: d2Vdx2=−3x\frac{d^2V}{dx^2} = -3x. At x=10x=10, d2Vdx2=−30\frac{d^2V}{dx^2} = -30, which is <0< 0, confirming a maximum.

Explanation:

To optimize the volume, we first reduce the volume formula to a single variable using the surface area constraint. By finding where the first derivative is zero, we locate the stationary point, and the second derivative test confirms it is the maximum capacity.