A-Level Edexcel Maths Pure Trigonometric Functions: 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 15 - 2020 - Paper 1

Question 15

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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 infl... 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 15 - 2020 - Paper 1

Step 1

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

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Answer

To differentiate the equation $x^2 \tan y = 9$ implicitly with respect to x, we apply the product rule:

Differentiate the left side: $2x \tan y + x^2 \sec^2 y \frac{dy}{dx} = 0$
Rearrange to isolate $\frac{dy}{dx}:$ $x^2 \sec^2 y \frac{dy}{dx} = -2x \tan y$ $\frac{dy}{dx} = \frac{-2x \tan y}{x^2 \sec^2 y}$
Simplify this using the identity: $\tan y = \frac{9}{x^2}$ and $\sec^2 y = 1 + \tan^2 y$ , where: $\sec^2 y = 1 + \left(\frac{9}{x^2}\right)^2 = \frac{x^4 + 81}{x^4}$ .
Therefore, we substitute back: $$\frac{dy}{dx} = \frac{-2x \cdot \frac{9}{x^2}}{x^2 \cdot \frac{x^4 + 81}{x^4}} = \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 find the point of inflection, we need to calculate the second derivative:

Starting from the first derivative: $\frac{dy}{dx} = \frac{-18x}{x^4 + 81}$
Use the quotient rule to differentiate: $\frac{d^2y}{dx^2} = \frac{(x^4 + 81)(-18) - (-18x)(4x^3)}{(x^4 + 81)^2}$ .
Simplifying gives: $\frac{d^2y}{dx^2} = \frac{-18(x^4 + 81) + 72x^4}{(x^4 + 81)^2} = \frac{54x^4 - 18(81)}{(x^4 + 81)^2}.$
For a point of inflection, set $\frac{d^2y}{dx^2} = 0$ : $54x^4 - 1458 = 0 \\ \Rightarrow x^4 = \frac{1458}{54} = 27 \\ \Rightarrow x = \sqrt{27}.$
Finally, check the signs of $\frac{d^2y}{dx^2}$ around $x = \sqrt{27}$ . The second derivative changes sign, confirming a point of inflection.

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 15 - 2020 - Paper 1

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 15 - 2020 - Paper 1

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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