A-Level Edexcel Maths Pure Vectors in 2 Dimensions: The curve C has equation $$x^2

The curve C has equation $$x^2 \tan y = 9$$ $$0 < y < \frac{\pi}{2}$$ (a) Show that $$\frac{dy}{dx} = \frac{-18x}{x^4 + 81}$$ (b) Prove that C has a point of inflection at $x = \sqrt{27}$. - Edexcel - A-Level Maths Pure - Question 2 - 2020 - Paper 2

Question 2

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The curve C has equation $$x^2 \tan y = 9$$ $$0 < y < \frac{\pi}{2}$$ (a) Show that $$\frac{dy}{dx} = \frac{-18x}{x^4 + 81}$$ (b) Prove that C has a point of in... show full transcript

Worked Solution & Example Answer:The curve C has equation $$x^2 \tan y = 9$$ $$0 < y < \frac{\pi}{2}$$ (a) Show that $$\frac{dy}{dx} = \frac{-18x}{x^4 + 81}$$ (b) Prove that C has a point of inflection at $x = \sqrt{27}$. - Edexcel - A-Level Maths Pure - Question 2 - 2020 - Paper 2

Step 1

Show that $ \frac{dy}{dx} = \frac{-18x}{x^4 + 81} $

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Answer

To find ( \frac{dy}{dx} ), we will differentiate the given equation implicitly.

Starting with the equation:

$x^2 \tan y = 9$

Differentiating both sides with respect to (x) gives:

$2x \tan y + x^2 \sec^2 y \frac{dy}{dx} = 0$

Rearranging, we obtain:

$x^2 \sec^2 y \frac{dy}{dx} = -2x \tan y$

Thus,

$\frac{dy}{dx} = \frac{-2x \tan y}{x^2 \sec^2 y}$

Utilizing the trigonometric identity ( \sec^2 y = 1 + \tan^2 y ), we can express ( \tan y ) in terms of ( x ). From the original equation:

$\tan y = \frac{9}{x^2}$

Then substituting this back in:

$\frac{dy}{dx} = \frac{-2x \left(\frac{9}{x^2}\right)}{x^2 \left(1 + \left(\frac{9}{x^2}\right)^2\right)}$

This simplifies to:

$\frac{dy}{dx} = \frac{-18}{x \left(1 + \frac{81}{x^4}\right)}$

Further simplifying gives:

$\frac{dy}{dx} = \frac{-18x}{x^4 + 81}$ .

Step 2

Prove that C has a point of inflection at $ x = \sqrt{27} $

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Answer

To determine the point of inflection, we need to find ( \frac{d^2y}{dx^2} ) and check its sign.

Starting from our earlier result:

$\frac{dy}{dx} = \frac{-18x}{x^4 + 81}$

Differentiating again using the quotient rule:

$\frac{d^2y}{dx^2} = \frac{(x^4 + 81)(-18) - (-18x)(4x^3)}{(x^4 + 81)^2}$

Simplifying this gives:

$\frac{d^2y}{dx^2} = \frac{-18(x^4 + 81) + 72x^4}{(x^4 + 81)^2}$

Further simplification results in:

$\frac{d^2y}{dx^2} = \frac{54x^4 - 18(81)}{(x^4 + 81)^2}$

Setting this equal to zero to find the point of inflection:

$54x^4 - 1458 = 0$

We find:

$x^4 = \frac{1458}{54} = 27 \implies x = \sqrt[4]{27} = \sqrt{3^3} = \sqrt{27}$ .

Now, we need to check the sign of ( \frac{d^2y}{dx^2} ) around ( x = \sqrt{27} ):

For ( x < \sqrt{27} ): ( \frac{d^2y}{dx^2} > 0 )
For ( x > \sqrt{27} ): ( \frac{d^2y}{dx^2} < 0 )

This change in sign indicates that there is a point of inflection at ( x = \sqrt{27} ).

The curve C has equation $$x^2 \tan y = 9$$ $$0 < y < \frac{\pi}{2}$$ (a) Show that $$\frac{dy}{dx} = \frac{-18x}{x^4 + 81}$$ (b) Prove that C has a point of inflection at $x = \sqrt{27}$. - Edexcel - A-Level Maths Pure - Question 2 - 2020 - Paper 2

Worked Solution & Example Answer:The curve C has equation $$x^2 \tan y = 9$$ $$0 < y < \frac{\pi}{2}$$ (a) Show that $$\frac{dy}{dx} = \frac{-18x}{x^4 + 81}$$ (b) Prove that C has a point of inflection at $x = \sqrt{27}$. - Edexcel - A-Level Maths Pure - Question 2 - 2020 - Paper 2

Show that \( \frac{dy}{dx} = \frac{-18x}{x^4 + 81} \)

Prove that C has a point of inflection at \( x = \sqrt{27} \)

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