The curve C is defined for $t \geq 0$ by the parametric equations $x = t^2 + t$ and $y = 4t^2 - t^3$ C is shown in the diagram below. Find the gradient of C at the point where it intersects the positive x-axis. --- The area A enclosed between C and the x-axis is given by $A = \int_0^b y \, dx$ Find the value of b. --- Use the substitution $y = 4t^2 - t^3$ to show that $A = \int_0^4 (4t^2 + 7t^1 - 2t^4) \, dt$ Find the value of A.

A-Level AQA Maths Pure Integration: The curve C is defined for $t \g

The curve C is defined for $t \geq 0$ by the parametric equations $x = t^2 + t$ and $y = 4t^2 - t^3$ C is shown in the diagram below - AQA - A-Level Maths Pure - Question 14 - 2021 - Paper 1

Question 14

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The curve C is defined for $t \geq 0$ by the parametric equations $x = t^2 + t$ and $y = 4t^2 - t^3$ C is shown in the diagram below. Find the gradient of C at th... show full transcript

Worked Solution & Example Answer:The curve C is defined for $t \geq 0$ by the parametric equations $x = t^2 + t$ and $y = 4t^2 - t^3$ C is shown in the diagram below - AQA - A-Level Maths Pure - Question 14 - 2021 - Paper 1

Step 1

Find the gradient of C at the point where it intersects the positive x-axis

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Answer

To find the point where the curve intersects the positive x-axis, we set the equation for y equal to zero:

4t^2 - t^3 = 0

Factoring gives:

4t^2 = t^3 \Rightarrow t(t^2 - 4) = 0

This results in t = 0 or t = 2. We take the positive 't', thus t = 2.

Now we compute the gradient at this point:

We calculate ( \frac{dy}{dt} ) and ( \frac{dx}{dt} ):
- $\frac{dy}{dt} = 8t - 3t^2$
- $\frac{dx}{dt} = 2t + 1$
At ( t = 2 ):
- $\frac{dy}{dt} = 8(2) - 3(2^2) = 16 - 12 = 4$
- $\frac{dx}{dt} = 2(2) + 1 = 4 + 1 = 5$
The gradient is then given by:

\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4}{5}.

Step 2

Find the value of b.

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Answer

Given the equations for x and y, we need to find the limits for the area calculation:

The point of intersection is where $y = 0$ , hence we use the previously calculated t values which intersect the x-axis.
Since we found t = 2 as the maximum point before it intersects back down at the x-axis, we can then determine that: $b = 2.$

Step 3

Use the substitution y = 4t^2 - t^3 to show that A = ∫₀² (4t² + 7t - 2t⁴) dt

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Answer

To show that:

Substitute ( y = 4t^2 - t^3 ): The bounds will remain at $0$ to $2$ since these are our t values.
Compute $\frac{dx}{dt}$ to adjust the area integral accordingly: $\frac{dx}{dt} = 2t + 1\implies dx = (2t+1)dt.$
Substitute into the integral for A:

A = \int_{0}^{2} (4t^2 - t^3)(2t + 1) dt = \int_{0}^{2} (4t^2(2t + 1) - t^3(2t + 1)) dt.

Expand and combine like terms: $A = \int_{0}^{2} (8t^3 + 4t^2 - 2t^4 - t^3) dt = \int_{0}^{2} (8t^3 + 4t^2 - 2t^4) dt.$

Step 4

Find the value of A.

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Answer

Integrate:

Using the simplified expression: $A = \int_{0}^{2} (6t^3 + 4t^2) dt$
Compute the integral: $A = \left[ \frac{6}{4}t^4 + \frac{4}{3}t^3 \right]_{0}^{2} = \left[ 6(4) + \frac{32}{3} \right] = 24 + \frac{32}{3} = \frac{72 + 32}{3} = \frac{104}{3}$

The curve C is defined for $t \geq 0$ by the parametric equations $x = t^2 + t$ and $y = 4t^2 - t^3$ C is shown in the diagram below - AQA - A-Level Maths Pure - Question 14 - 2021 - Paper 1

Worked Solution & Example Answer:The curve C is defined for $t \geq 0$ by the parametric equations $x = t^2 + t$ and $y = 4t^2 - t^3$ C is shown in the diagram below - AQA - A-Level Maths Pure - Question 14 - 2021 - Paper 1

Find the gradient of C at the point where it intersects the positive x-axis

Find the value of b.

Use the substitution y = 4t^2 - t^3 to show that A = ∫₀² (4t² + 7t - 2t⁴) dt

Find the value of A.

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