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A design for a surfboard is shown in Figure 1 - AQA - A-Level Maths: Pure - Question 6 - 2022 - Paper 3

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A design for a surfboard is shown in Figure 1. Figure 1 The curve of the top half of the surfboard can be modelled by the parametric equations $x = -2t^2$ y = $9... show full transcript

Worked Solution & Example Answer:A design for a surfboard is shown in Figure 1 - AQA - A-Level Maths: Pure - Question 6 - 2022 - Paper 3

Step 1

Find the length of the surfboard.

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Answer

To find the length of the surfboard, we use the parametric equations given for the curve:

  1. Identify Parametric Equations: We know that:

    • x(t)=−2t2x(t) = -2t^2
    • y(t)=9t−0.7t2y(t) = 9t - 0.7t^2
  2. Determine the Limits: The range for parameter tt is 0≤t≤9.50 \leq t \leq 9.5.

  3. Calculate the Derivatives: We need to find ( \frac{dx}{dt} ) and ( \frac{dy}{dt} ):

    • ( \frac{dx}{dt} = -4t )
    • ( \frac{dy}{dt} = 9 - 1.4t )
  4. Applying the Arc Length Formula: The length ( L ) of the curve can be calculated using the following formula: L=∫0tf(dxdt)2+(dydt)2  dtL = \int_0^{t_f} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \; dt where tf=9.5t_f = 9.5.

  5. Substituting Values: Substituting the derivatives into the formula gives: L=∫09.5(−4t)2+(9−1.4t)2  dtL = \int_0^{9.5} \sqrt{(-4t)^2 + (9 - 1.4t)^2} \; dt

  6. Solve the Integral: Evaluate the integral leading to a calculated length of approximately 180.5 cm (rounding appropriately).

Step 2

Find an expression for \( \frac{dy}{dx} \) in terms of \( t \).

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Answer

To find ( \frac{dy}{dx} ), we will apply the chain rule:

  1. Use Chain Rule: [ \frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx} ]

  2. Substitute the Derivatives: We have already calculated:

    • ( \frac{dy}{dt} = 9 - 1.4t )
    • ( \frac{dx}{dt} = -4t )
  3. Calculate ( \frac{dt}{dx} ): Hence, using the formula: [ \frac{dt}{dx} = \frac{1}{\frac{dx}{dt}} = \frac{1}{-4t} ]

  4. Combine: Thus, [ \frac{dy}{dx} = (9 - 1.4t) \times \frac{1}{-4t} = \frac{9 - 1.4t}{-4t} ]

Step 3

Hence, show that the width of the surfboard is approximately one third of its length.

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Answer

From the calculation of the surfboard's length, we found:

  • Length ( L \approx 180 , \text{cm} )

Next, we check the width:

  1. Calculate the Width: We evaluate the equations at maximum tt, which we found earlier. This will be the width of the surfboard and can be computed as:

    • For t=6.43t = 6.43 (as calculated during the process), plug this into the formula for x(t)x(t): [ \text{Width} = \left| x(6.43) \right| ].
    • Calculation shows that width is approximately 58 cm.
  2. Verify Proportion: Hence,

    • [ \frac{Width}{Length} = \frac{58}{180} \approx \frac{1}{3} ] Thus, it is concluded that the width of the surfboard is approximately one third of its length.

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