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Calculus - Applications of calculus

Grade 11IB_AI

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

🔑Concepts

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The derivative f′(x)f'(x) represents the gradient of the tangent to the curve y=f(x)y = f(x) at any point xx.

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A tangent to a curve at a point is a straight line that just touches the curve at that point. Its gradient is m=f′(a)m = f'(a).

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A normal to a curve at a point is a straight line perpendicular to the tangent at that point. Its gradient is m⊥=−1f′(a)m_{\perp} = -\frac{1}{f'(a)}.

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Local maximum and minimum points (stationary points) occur where the first derivative is zero, i.e., f′(x)=0f'(x) = 0.

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Optimization involves finding the maximum or minimum value of a function within a given context, such as area, volume, or cost.

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The Trapezoidal Rule is used to approximate the area under a curve when the function is not easily integrable or is given as a set of data points.

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In kinematics, if s(t)s(t) is the displacement, then velocity v(t)=s′(t)v(t) = s'(t) and acceleration a(t)=v′(t)=s′′(t)a(t) = v'(t) = s''(t).

📐Formulae

mtangent=f′(a)m_{tangent} = f'(a) accounts for the gradient at x=ax=a

y−y1=m(x−x1)y - y_1 = m(x - x_1) (Equation of a straight line)

mnormal=−1f′(a)m_{normal} = -\frac{1}{f'(a)}

A≈12h[(y0+yn)+2(y1+y2+...+yn−1)]A \approx \frac{1}{2}h [ (y_0 + y_n) + 2(y_1 + y_2 + ... + y_{n-1}) ] where h=b−anh = \frac{b-a}{n}

f′(x)=0f'(x) = 0 (Condition for stationary points)

💡Examples

Problem 1:

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

Solution:

  1. Find the y-coordinate: f(2)=2(2)2−3(2)+1=8−6+1=3f(2) = 2(2)^2 - 3(2) + 1 = 8 - 6 + 1 = 3. The point is (2,3)(2, 3).
  2. Find the derivative: f′(x)=4x−3f'(x) = 4x - 3.
  3. Find the gradient at x=2x = 2: m=f′(2)=4(2)−3=5m = f'(2) = 4(2) - 3 = 5.
  4. Use the point-slope form: y−3=5(x−2)y - 3 = 5(x - 2).
  5. Simplify: y=5x−10+3⇒y=5x−7y = 5x - 10 + 3 \Rightarrow y = 5x - 7.

Explanation:

We first find the coordinates of the point, then use the derivative to find the slope of the tangent, and finally apply the linear equation formula.

Problem 2:

A rectangular garden is to be fenced against a straight wall. If 40 meters of fencing are available for the other three sides, find the dimensions that maximize the area.

Solution:

  1. Let xx be the width perpendicular to the wall and yy be the length parallel to the wall.
  2. Constraint: 2x+y=40⇒y=40−2x2x + y = 40 \Rightarrow y = 40 - 2x.
  3. Area function: A=x⋅y=x(40−2x)=40x−2x2A = x \cdot y = x(40 - 2x) = 40x - 2x^2.
  4. Differentiate: A′(x)=40−4xA'(x) = 40 - 4x.
  5. Set A′(x)=0A'(x) = 0: 40−4x=0⇒x=1040 - 4x = 0 \Rightarrow x = 10.
  6. Find yy: y=40−2(10)=20y = 40 - 2(10) = 20.
  7. Dimensions: 1010 m by 2020 m.

Explanation:

To maximize area, we express the area in terms of one variable using the perimeter constraint, then find where the derivative of the area function equals zero.

Problem 3:

Use the trapezoidal rule with n=2n=2 intervals to estimate the area under f(x)=x2f(x) = x^2 from x=0x=0 to x=2x=2.

Solution:

  1. Calculate step size: h=2−02=1h = \frac{2 - 0}{2} = 1.
  2. Determine x-values: x0=0,x1=1,x2=2x_0 = 0, x_1 = 1, x_2 = 2.
  3. Calculate y-values: y0=02=0y_0 = 0^2 = 0, y1=12=1y_1 = 1^2 = 1, y2=22=4y_2 = 2^2 = 4.
  4. Apply formula: Area≈12(1)[(0+4)+2(1)]Area \approx \frac{1}{2}(1) [ (0 + 4) + 2(1) ].
  5. Area≈0.5[4+2]=0.5×6=3Area \approx 0.5 [ 4 + 2 ] = 0.5 \times 6 = 3.

Explanation:

The trapezoidal rule sums the areas of two trapezoids created under the curve using the specified interval width.

Applications of calculus Grade 11 Notes & Examples