A-Level Edexcel Maths Pure Integration: 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 1 - 2020 - Paper 1

Question 1

$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 1-2020-Paper 1.png$

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 1 - 2020 - Paper 1

Step 1

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

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Answer

To find $\frac{dy}{dx}$ , we'll start by differentiating the given equation implicitly.

Differentiate both sides of the equation $x^2 \tan y = 9$ with respect to $x$ : $\frac{d}{dx}(x^2 \tan y) = 0$ .
Using the product rule, we have: $2x \tan y + x^2 \sec^2 y \frac{dy}{dx} = 0$ .
Rearranging gives us: $x^2 \sec^2 y \frac{dy}{dx} = -2x \tan y$ .
Thus, $\frac{dy}{dx} = \frac{-2x \tan y}{x^2 \sec^2 y}$ .
Recognizing that $\tan y = \frac{9}{x^2}$ from the original equation, we substitute: $\sec^2 y = 1 + \tan^2 y = 1 + \left(\frac{9}{x^2}\right)^2 = \frac{x^4 + 81}{x^4}$ .
Substituting back in gives: $\frac{dy}{dx} = \frac{-2x \cdot \frac{9}{x^2}}{x^2 \cdot \frac{x^4 + 81}{x^4}} = \frac{-18x}{x^4 + 81}.$

Step 2

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

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Answer

To prove that C has a point of inflection at $x = \sqrt{27}$ , we need to examine the second derivative:

Start from the expression we derived for $\frac{dy}{dx}$ : $\frac{dy}{dx} = \frac{-18x}{x^4 + 81}.$
Differentiate 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 us: $\frac{d^2y}{dx^2} = \frac{-18(x^4 + 81) + 72x^4}{(x^4 + 81)^2} = \frac{54x^4 - 18\cdot 81}{(x^4 + 81)^2}.$
Setting $x = \sqrt{27}$ , we calculate:
- First, calculate $54(\sqrt{27})^4 - 18 \cdot 81$ , which equals zero.
Since $\frac{d^2y}{dx^2}$ changes sign around $x = \sqrt{27}$ , this confirms that there is a point of inflection.

Thus, C has 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 1 - 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 1 - 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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