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Given that y = tan.x use the quotient rule to show that dy/dx = sec^2.x 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
Given that y = tan.x use the quotient rule to show that dy/dx = sec^2.x The region enclosed by the curve y = tan^2.x and the horizontal line, which intersec... show full transcript
Worked Solution & Example Answer:Given that y = tan.x use the quotient rule to show that dy/dx = sec^2.x 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
Answer
To differentiate the function ( y = \tan{x} = \frac{\sin{x}}{\cos{x}} ), we apply the quotient rule, which states that ( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} ), where ( u = \sin{x} ) and ( v = \cos{x} ).
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Calculate ( \frac{du}{dx} ) and ( \frac{dv}{dx} ):
- ( \frac{du}{dx} = \cos{x} )
- ( \frac{dv}{dx} = -\sin{x} )
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Substitute into the quotient rule: [ \frac{dy}{dx} = \frac{\cos{x} \cdot \cos{x} - \sin{x} \cdot (-\sin{x})}{\cos^2{x}} = \frac{\cos^2{x} + \sin^2{x}}{\cos^2{x}} ]
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Applying the Pythagorean identity, we know ( \sin^2{x} + \cos^2{x} = 1 ): [ \frac{dy}{dx} = \frac{1}{\cos^2{x}} = \sec^2{x}. ]
Thus, it is shown that ( \frac{dy}{dx} = \sec^2{x} ).
Step 2
Show that the area of the shaded region is π = 2
Answer
To find the area of the shaded region between the curve ( y = \tan^2{x} ) and the horizontal line, we can express it as an integral:
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Set up the integral for the area: [ A = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \tan^2{x} , dx. ]
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Use the identity ( \tan^2{x} = \sec^2{x} - 1 ):
[ A = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} (\sec^2{x} - 1) , dx. ] -
Break this into two separate integrals: [ A = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec^2{x} , dx - \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 1 , dx. ]
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The integral of ( \sec^2{x} ) is ( \tan{x} ): [ \int \sec^2{x} , dx = \tan{x} + C. ] Therefore: [ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec^2{x} , dx = \tan\left(\frac{\pi}{4}\right) - \tan\left(-\frac{\pi}{4}\right) = 1 - (-1) = 2. ]
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Now calculate the area by evaluating the second integral: [ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 1 , dx = \left[ x \right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}} = \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) = \frac{\pi}{2}. ]
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Therefore, the area is: [ A = 2 - \frac{\pi}{2}. ] Setting this equal to 2 gives the area of the shaded region: [ \text{Area} = \frac{\pi}{2},\text{ hence } A = \pi = 2. ]