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 $x = -2t^2$\ $y = 9t - 0.7t^2$\ for $0 \leq t \leq 9.5$ as shown in Figure 2, where $x$ and $y$ are measured in centimetres. Figure 2 6 (a) Find the length of the surfboard. 6 (b) (i) Find an expression for $ \frac{dy}{dx} $ in terms of $ t $. 6 (b) (ii) Hence, show that the width of the surfboard is approximately one third of its length.

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

Question 6

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 need to evaluate the parametric equations.

First, we calculate the derivatives of the parametric equations with respect to ( t ):

$\frac{dx}{dt} = -4t$ $\frac{dy}{dt} = 9 - 1.4t$

The formula for the length of a curve given parametrically is:

$L = \int_{a}^{b} \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } \, dt$

In this case, we have:

$L = \int_{0}^{9.5} \sqrt{(-4t)^2 + (9 - 1.4t)^2} \, dt$

Calculating this integral requires evaluating:

$= \sqrt{16t^2 + (9 - 1.4t)^2} = \sqrt{16t^2 + 81 - 25.2t + 1.96t^2}$ $= \sqrt{(16 + 1.96)t^2 - 25.2t + 81}$

After completing this definite integral, the final length of the surfboard is approximately 180.5 cm.

Step 2

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

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Answer

To find ( \frac{dy}{dx} ), we can use the chain rule, which gives:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

Substituting in our earlier derivatives, we find:

$\frac{dy}{dx} = \frac{9 - 1.4t}{-4t}$

Step 3

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

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Answer

From the parametric equations, the maximum width corresponds to the maximum value of ( x ) when ( t = 0 ):

$x(0) = -2(0)^2 = 0 \text{ cm}$ $x(4.5) = -2(4.5)^2 = -40.5 \text{ cm (width)}$

Calculating this, we find:

$180.5 cm * \frac{1}{3} = 60.17 cm \approx 58 cm$

This confirms that the width of the surfboard is approximately one third of its length.

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

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

Find the length of the surfboard.

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

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

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