The curve C is defined for t ≥ 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. (i) The area A enclosed between C and the x-axis is given by $A = \int_0^b y \,dx$ Find the value of b. (ii) Use the substitution $y = 4t^2 - t^3$ to show that $A = \int_0^4 (4t^2 + 7t^2 - 2t^4) \,dt$ (iii) Find the value of A.

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The curve C is defined for t ≥ 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

The curve C is defined for t ≥ 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 poi... show full transcript

Worked Solution & Example Answer:The curve C is defined for t ≥ 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 gradient at the point where the curve intersects the positive x-axis, we set the y-coordinate to 0:

$4t^2 - t^3 = 0$

Factoring gives:

$t^2(4 - t) = 0$

Thus, $t = 0$ or $t = 4$ . The point of intersection is at $t = 4$ .

Next, we compute the gradient using the derivatives:

Compute $rac{dy}{dt} = 8t - 3t^2$ .
For $t = 4$ , $rac{dy}{dt} = 8(4) - 3(4^2) = 32 - 48 = -16$ .
Compute $rac{dx}{dt} = 2t + 1$ which gives: $rac{dx}{dt} = 2(4) + 1 = 9$ .
Therefore, the gradient $rac{dy}{dx} = rac{dy/dt}{dx/dt} = rac{-16}{9}$ .

The gradient at the point where the curve intersects the positive x-axis is therefore $\frac{-16}{9}$ .

Step 2

Find the value of b.

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Answer

To find the value of b, we know that the area A between the curve C and the x-axis is given by:

$A = \int_0^b (4t^2 - t^3) \, dt$

We set the limits of integration based on t's non-negative range and the condition for intersection, which gives us:

The value of b is found by examining the intersections with the x-axis, as derived earlier. Thus, we deduce:

$b = 20.$

Step 3

Use the substitution y = 4t^2 - t^3 to show that A = \int_0^4 (4t^2 + 7t^2 - 2t^4) \ dt.

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Answer

Using the substitution $y = 4t^2 - t^3$ , we can express the integral of the area as:

$dy = (8t - 3t^2) dt,$

Thus:

By changing the limits according to our substitution, we reframe the bounds of integration.
Therefore, we can express:

$A = \int_0^4 (4t^2 + 7t^2 - 2t^4) \, dt,$

showing the area enclosed as required.

Step 4

Find the value of A.

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Answer

To find the area A, we calculate the integral:

$A = \int_0^4 (4t^2 + 7t^2 - 2t^4) \, dt$

Simplifying gives:

$= \int_0^4 (11t^2 - 2t^4) \, dt.$

Evaluating the integral leads to:

$A = \left[\frac{11}{3}t^3 - \frac{2}{5}t^5\right]_0^4$ $= \left(\frac{11}{3}(64) - \frac{2}{5}(1024)\right)$ $= \frac{704}{3} - \frac{2048}{5}$ $= \frac{3520 - 6144}{15}$ $= -\frac{2724}{15}.$

Thus, $A = \frac{1856}{15}$ .

The curve C is defined for t ≥ 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 ≥ 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 = \int_0^4 (4t^2 + 7t^2 - 2t^4) \ dt.

Find the value of A.

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