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5. (i) Find, using calculus, the x coordinate of the turning point of the curve with equation y = e^x cos 4x, \frac{\pi}{4} < x < \frac{\pi}{2} Give your answer to 4 decimal places - Edexcel - A-Level Maths Pure - Question 7 - 2016 - Paper 3
Question 7
5. (i) Find, using calculus, the x coordinate of the turning point of the curve with equation y = e^x cos 4x, \frac{\pi}{4} < x < \frac{\pi}{2} ... show full transcript
Worked Solution & Example Answer:5. (i) Find, using calculus, the x coordinate of the turning point of the curve with equation y = e^x cos 4x, \frac{\pi}{4} < x < \frac{\pi}{2} Give your answer to 4 decimal places - Edexcel - A-Level Maths Pure - Question 7 - 2016 - Paper 3
Step 1
Find, using calculus, the x coordinate of the turning point of the curve with equation y = e^x cos 4x, \frac{\pi}{4} < x < \frac{\pi}{2}
Answer
To find the x-coordinate of the turning point, we need to calculate the derivative of the function and set it to zero.
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Differentiate y with respect to x:
\frac{dy}{dx} = e^x (cos 4x - 4sin 4x)
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Set the derivative to zero for turning points:
e^x (cos 4x - 4sin 4x) = 0
Since e^x is never zero, we focus on:
cos 4x - 4sin 4x = 0
=> cos 4x = 4sin 4x
=> tan 4x = \frac{1}{4}
The solution for 4x can be found using the arctangent function:
4x = arctan(\frac{1}{4}) + k\pi, \ k \in \mathbb{Z}
To find the valid values of x, we can calculate:
x = \frac{1}{4} arctan(\frac{1}{4}) + \frac{k\pi}{4}
Analyze the bounds \frac{\pi}{4} < x < \frac{\pi}{2} to find k values that fit.
Evaluate this numerically to find x = 0.9463.
Step 2
Given x = sin^2 2y, 0 < y < \frac{\pi}{4}, find \frac{dy}{dx} as a function of y.
Answer
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First, differentiate x with respect to y:
\frac{dx}{dy} = 2sin(2y)cos(2y) = sin(4y)
- Hence, using the relationship:
\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{1}{sin(4y)}
- Using the identity, we can rewrite this as:
\frac{dy}{dx} = \frac{1}{2}cosec(2y)
with the necessary constants p and q as follows:
where p = \frac{1}{2}, q = 2.