A-Level AQA Maths Pure Differentiation: 10 (a) Given that y = tan.x u

10 (a) Given that y = tan.x use the quotient rule to show that dy/dx = sec^2.x 10 (b) The region enclosed by the curve y = tan^2.x and the horizontal line, which intersects the curve at x = -π/4 and x = π/4 is shaded in the diagram below - AQA - A-Level Maths Pure - Question 10 - 2021 - Paper 1

Question 10

10-(a)-Given-that---y-=-tan.x--use-the-quotient-rule-to-show-that---dy/dx-=-sec^2.x--10-(b)-The-region-enclosed-by-the-curve-y-=-tan^2.x-and-the-horizontal-line,-which-intersects-the-curve-at-x-=--π/4-and-x-=-π/4-is-shaded-in-the-diagram-below-AQA-A-Level Maths Pure-Question 10-2021-Paper 1.png

10 (a) Given that y = tan.x use the quotient rule to show that dy/dx = sec^2.x 10 (b) The region enclosed by the curve y = tan^2.x and the horizontal line, whi... show full transcript

Worked Solution & Example Answer:10 (a) Given that y = tan.x use the quotient rule to show that dy/dx = sec^2.x 10 (b) The region enclosed by the curve y = tan^2.x and the horizontal line, which intersects the curve at x = -π/4 and x = π/4 is shaded in the diagram below - AQA - A-Level Maths Pure - Question 10 - 2021 - Paper 1

Step 1

Given that y = tan.x, use the quotient rule to show that dy/dx = sec^2.x

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Answer

To differentiate the function, we start with:

$y = rac{ ext{sin}(x)}{ ext{cos}(x)}$ .

Using the quotient rule:

$rac{dy}{dx} = rac{( ext{cos}(x) imes ext{d(sin}(x))/dx) - ( ext{sin}(x) imes ext{d(cos}(x))/dx)}{ ext{cos}^2(x)}$ .

We know:

$rac{d( ext{sin}(x))}{dx} = ext{cos}(x)$
$rac{d( ext{cos}(x))}{dx} = - ext{sin}(x)$

Substituting these derivatives gives:

$rac{dy}{dx} = rac{( ext{cos}(x) imes ext{cos}(x)) - ( ext{sin}(x) imes - ext{sin}(x))}{ ext{cos}^2(x)}$

This simplifies to:

$rac{dy}{dx} = rac{ ext{cos}^2(x) + ext{sin}^2(x)}{ ext{cos}^2(x)}$

Using the Pythagorean identity, $ext{sin}^2(x) + ext{cos}^2(x) = 1$ , we find that:

$rac{dy}{dx} = rac{1}{ ext{cos}^2(x)}$

Thus,

$rac{dy}{dx} = ext{sec}^2(x)$ .

Step 2

Show that the area of the shaded region is π/2

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Answer

To find the area of the shaded region, we will set up an integral from the points of intersection:

$A = ext{Area} = \\int_{-π/4}^{π/4} ext{tan}^2(x) \, dx$ .

To integrate this, we can use the identity:

$ext{tan}^2(x) = ext{sec}^2(x) - 1$ .

Thus, the integral becomes:

$A = \\int_{-π/4}^{π/4} ( ext{sec}^2(x) - 1) \, dx$ .

This can be split into two integrals:

$A = \\int_{-π/4}^{π/4} ext{sec}^2(x) \, dx - \\int_{-π/4}^{π/4} 1 \, dx$ .

Calculating the first integral:

$\int ext{sec}^2(x) \, dx = ext{tan}(x)$ ,

$A = [ ext{tan}(x)]_{-π/4}^{π/4} - [-x]_{-π/4}^{π/4}$ .

Evaluating,

$A = [ ext{tan}(π/4) - ext{tan}(-π/4)] - [π/4 - (-π/4)] = [1 - (-1)] - [π/2]$ .

Thus,

$A = 2 - rac{π}{2}$ .

But, we notice that the shaded area is indeed half of this due to the symmetry, which brings us to:

$A = rac{π}{2}$ .

Hence, the area of the shaded region is confirmed to be:

$rac{π}{2}$ .

10 (a) Given that y = tan.x use the quotient rule to show that dy/dx = sec^2.x 10 (b) The region enclosed by the curve y = tan^2.x and the horizontal line, which intersects the curve at x = -π/4 and x = π/4 is shaded in the diagram below - AQA - A-Level Maths Pure - Question 10 - 2021 - Paper 1

Given that y = tan.x, use the quotient rule to show that dy/dx = sec^2.x

Show that the area of the shaded region is π/2

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