The curve C is defined for t ≥ 0 by the parametric equations x = t² + t and y = 4t² - t³ 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 = ∫[0 to b] y dx Find the value of b. (ii) Use the substitution y = 4t² - t³ to show that A = ∫[0 to 4] (4t² + 7t - 2t⁴) dt (iii) Find the value of A.

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The curve C is defined for t ≥ 0 by the parametric equations x = t² + t and y = 4t² - t³ C is shown in the diagram below - AQA - A-Level Maths: Mechanics - Question 14 - 2021 - Paper 1

Question 14

The curve C is defined for t ≥ 0 by the parametric equations x = t² + t and y = 4t² - t³ C is shown in the diagram below. Find the gradient of C at the poin... show full transcript

Worked Solution & Example Answer:The curve C is defined for t ≥ 0 by the parametric equations x = t² + t and y = 4t² - t³ C is shown in the diagram below - AQA - A-Level Maths: Mechanics - 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 y = 0:

4t^2 - t^3 = 0

Factoring gives:

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

This gives t = 0 or t = 4. We take t = 4 because we are interested in the positive intersection.

Now we find the x-coordinate corresponding to t = 4:

x = 4^2 + 4 = 16 + 4 = 20

Next, we need to find the gradient of the curve at this point using the formula:

rac{dy}{dx} = rac{ rac{dy}{dt}}{ rac{dx}{dt}}

Calculating the derivatives:

For y: $rac{dy}{dt} = 8t - 3t^2$
Substituting t = 4 gives: $rac{dy}{dt} = 8(4) - 3(4)^2 = 32 - 48 = -16$
For x: $rac{dx}{dt} = 2t + 1$ Substituting t = 4 gives: $rac{dx}{dt} = 2(4) + 1 = 8 + 1 = 9$

Now, substituting these results into the gradient formula:

$rac{dy}{dx} = rac{-16}{9}$

Step 2

Find the value of b.

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Answer

The area A is given by:

A = ext{∫ [0 to b]} y \, dx

From the equations:

$dx = (2t + 1)dt$ $y = 4t^2 - t^3$

Setting the limits for x from t = 0 to t = b, we need the relationship between x and t:

Setting the equation equal to b gives:

$b = t^2 + t\ ext{, } t=0 ext{ gives } b=0\ ext{, at } t=4 \ ext{ gives } b = 20$ .

Step 3

Use the substitution y = 4t² - t³ to show that A = ∫[0 to 4] (4t² + 7t - 2t⁴) dt.

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Answer

Using the substitution for y, we need to calculate A:

Substituting for y gives:

A = ext{∫ [0 to 4]} (4t^2 - t^3)(2t + 1) dt = ext{∫ [0 to 4]} (4t^2 + 7t - 2t^4) dt

This confirms the relationship required.

Step 4

Find the value of A.

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Answer

To find A, we evaluate:

A = ext{∫ [0 to 4]} (4t^2 + 7t - 2t^4) dt

Calculating this:

Integral of 4t² is: $\frac{4t^3}{3}$ for t from 0 to 4.
Integral of 7t is: $\frac{7t^2}{2}$ for t from 0 to 4.
Integral of -2t⁴ is: $\frac{-2t^5}{5}$ for t from 0 to 4.

Evaluating at t = 4, we get: $rac{4(4^3)}{3} + rac{7(4^2)}{2} - rac{2(4^5)}{5}$ Thus, $A = rac{1856}{15}$

The curve C is defined for t ≥ 0 by the parametric equations x = t² + t and y = 4t² - t³ C is shown in the diagram below - AQA - A-Level Maths: Mechanics - Question 14 - 2021 - Paper 1

Worked Solution & Example Answer:The curve C is defined for t ≥ 0 by the parametric equations x = t² + t and y = 4t² - t³ C is shown in the diagram below - AQA - A-Level Maths: Mechanics - 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 = 4t² - t³ to show that A = ∫[0 to 4] (4t² + 7t - 2t⁴) dt.

Find the value of A.

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