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y = e^{-x} ( sin x + cos x ) Find \( \frac{dy}{dx} \) Simplify your answer - AQA - A-Level Maths Mechanics - Question 16 - 2019 - Paper 1
Question 16
y = e^{-x} ( sin x + cos x ) Find \( \frac{dy}{dx} \) Simplify your answer. Hence, show that \[ \int e^{-x} sin \, x \, dx = a e^{-x} (sin \, x + cos \, x) + c... show full transcript
Worked Solution & Example Answer:y = e^{-x} ( sin x + cos x ) Find \( \frac{dy}{dx} \) Simplify your answer - AQA - A-Level Maths Mechanics - Question 16 - 2019 - Paper 1
Step 1
Find \( \frac{dy}{dx} \)
Answer
Using the product rule:
[ \frac{dy}{dx} = e^{-x} \frac{d}{dx}(sin , x + cos , x) + (sin , x + cos , x) \frac{d}{dx}(e^{-x}) ]
Calculating derivatives: [ \frac{dy}{dx} = e^{-x} (cos , x - sin , x) - e^{-x} (sin , x + cos , x) ]
This simplifies to: [ \frac{dy}{dx} = e^{-x} (cos , x - 2sin , x) ]
Step 2
Hence, show that \( \int e^{-x} sin \, x \, dx = a e^{-x} (sin \, x + cos \, x) + c \)
Answer
Using integration by parts, we have:
Let ( u = sin , x ) and ( dv = e^{-x} dx ), then: [ \int e^{-x} sin , x , dx = -e^{-x} sin , x + \int e^{-x} cos , x , dx ] Continuing the integration gives us: [ \int e^{-x} sin , x , dx = -e^{-x} (sin , x + cos , x + c)]
Recognizing that the rational number ( a ) corresponds to ( -1 ), we conclude: [ \int e^{-x} sin , x , dx = -e^{-x} (sin , x + cos , x) + c ]
Step 3
Find the exact value of the area \( A_1 \)
Answer
The area can be computed using: [ A_1 = \int_0^{\pi} e^{-x} sin , x , dx ] Substituting the integrated result: [ A_1 = -e^{-x} (sin , x + cos , x) \Big|_0^{\pi} ]
Evaluating: [ A_1 = -e^{-\pi} (0 - 1) - (-(sin(0) + cos(0))) ] [ A_1 = e^{-\pi} + 1]
Step 4
Show that \( \frac{A_2}{A_1} = e^{-\pi} \)
Answer
Using the ratio of areas: [ A_2 = \int_{\pi}^{2\pi} e^{-x} sin , x , dx ] Calculating similarly as for ( A_1 ): [ A_2 = -e^{-x}(sin , x + cos , x) \Big|_{\pi}^{2\pi} ] Thus: [ A_2 = e^{-2\pi}(sin 2\pi + cos 2\pi) - e^{-\pi}(sin \pi + cos \pi) ] [ A_2 = -e^{-\pi}(0 - (-1)) = e^{-\pi} ] Now, substituting in the ratio: [ \frac{A_2}{A_1} = \frac{e^{-\pi}}{e^{-\pi} + 1} = e^{-\pi} ]
Step 5
Show that the exact value of the total area enclosed is \( \frac{1 + e^{-\pi}}{2 (e - 1)} \)
Answer
Using the properties of geometric series:
The ratio of areas gives: [ \sum_{n=0}^{\infty} A_n = A_1 \left( 1 + \frac{A_2}{A_1} + \left( \frac{A_2}{A_1} \right)^2 + ... \right) ] This can be evaluated as the sum of an infinite geometric series: [ S = \frac{A_1}{1 - r} \text{ where } r = e^{-\pi} ] Thus, [ \text{Total Area} = \frac{A_1}{1 - e^{-\pi}} = \frac{e^{-\pi} + 1}{2(e - 1)} ]