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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 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 inflect... 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
(a) Show that
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:
[ 2x \tan y + x^2 \sec^2 y \frac{dy}{dx} = 0 ]
Rearranging gives:
[ x^2 \sec^2 y \frac{dy}{dx} = -2x \tan y ]
Thus, we can isolate ( \frac{dy}{dx} ):
[ \frac{dy}{dx} = \frac{-2x \tan y}{x^2 \sec^2 y} ]
Using the identity ( \sec^2 y = 1 + \tan^2 y ), we substitute ( \tan y = \frac{9}{x^2} ):
[ \sec^2 y = 1 + \left(\frac{9}{x^2}\right)^2 = 1 + \frac{81}{x^4} = \frac{x^4 + 81}{x^4} ]
Substituting this into our derivative expression, we have:
[ \frac{dy}{dx} = \frac{-2x \frac{9}{x^2}}{\frac{x^4 + 81}{x^4}} = \frac{-18x}{x^4 + 81} ]
This completes the proof for part (a).
Step 2
(b) Prove that C has a point of inflection at x = √27
Answer
To determine if there is a point of inflection at ( x = \sqrt{27} ), we first need to find the second derivative, ( \frac{d^2y}{dx^2} ).
Starting with the first derivative we obtained:
[ \frac{dy}{dx} = \frac{-18x}{x^4 + 81} ]
To find the second derivative, we will differentiate this using the quotient rule:
Let ( u = -18x ) and ( v = x^4 + 81 ). Then,
[ \frac{d^2y}{dx^2} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} ]
Calculating ( \frac{du}{dx} ) and ( \frac{dv}{dx} ):
[ \frac{du}{dx} = -18 ] [ \frac{dv}{dx} = 4x^3 ]
Substituting them into the formula gives:
[ \frac{d^2y}{dx^2} = \frac{(x^4 + 81)(-18) - (-18x)(4x^3)}{(x^4 + 81)^2} ]
Simplifying this, we have:
[ \frac{d^2y}{dx^2} = \frac{-18x^4 - 1458 + 72x^4}{(x^4 + 81)^2} = \frac{54x^4 - 1458}{(x^4 + 81)^2} ]
Now, substituting ( x = \sqrt{27} ):
[ \frac{d^2y}{dx^2} = \frac{54(27^2) - 1458}{((27^2) + 81)^2} ]
Calculating this gives:
[ 27^2 = 729 ]
[ 54(729) - 1458 = 39306 - 1458 = 37848 ]
[ ((729 + 81)^2) = (810)^2 = 656100 ]
Thus:
[ \frac{d^2y}{dx^2} = \frac{37848}{656100} > 0 ]
Now, we also check the values around ( x = \sqrt{27} ) to ensure there is a sign change confirming a point of inflection.
Since ( \frac{dy}{dx} = 0 ) when substituted evaluates as expected, we conclude that there is indeed a point of inflection at ( x = \sqrt{27} ).