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. (ii) Given $x = \sin^2 2y$, $0 < y < \frac{\pi}{4}$, find $\frac{dy}{dx}$ as a function of $y$. Write your answer in the form $\frac{dy}{dx} = p \csc(qy)$, where $p$ and $q$ are constants to be determined.

A-Level Edexcel Maths Pure Equation of a Straight Line: 5. (i) Find, using calculus, the

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 6 - 2016 - Paper 3

Question 6

$5.-(i)-Find,-using-calculus,-the-x-coordinate-of-the-turning-point-of-the-curve-with-equation--------$y-=-e^x-\cos-4x$,---$\frac{\pi}{4}-<-x-<-\frac{\pi}{2}$----------Give-your-answer-to-4-decimal-places-Edexcel-A-Level Maths Pure-Question 6-2016-Paper 3.png$

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 6 - 2016 - Paper 3

Step 1

Find the turning point of the curve with equation $y = e^x \cos 4x$

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Answer

To find the turning point, we need to calculate the derivative of the function:

$\frac{dy}{dx} = e^x \cos 4x - 4e^x \sin 4x$

Set the derivative to zero:

$e^x \cos 4x - 4e^x \sin 4x = 0$

This simplifies to:

$\cos 4x = 4 \sin 4x$

$\tan 4x = \frac{1}{4}$

Now, find the value of $4x$ :

$4x = \tan^{-1}\left(\frac{1}{4}\right)$

Thus,

$x = \frac{1}{4} \tan^{-1}\left(\frac{1}{4}\right)$

Evaluating $x$ gives approximately:

$x \approx 0.2363$

So the x-coordinate of the turning point is $0.2363$ (to 4 decimal places).

Step 2

Given $x = \sin^2 2y$, find $\frac{dy}{dx}$ as a function of $y$

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Answer

Using the chain rule:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

where $u = \sin 2y$ .

First, find $\frac{du}{dx}$ :

$u = \sin 2y \Rightarrow \frac{du}{dx} = 2\sin 2y \cos 2y$

Now, differentiate $u$ with respect to $y$ :

$\frac{du}{dy} = 2\cos 2y$

Therefore,

$\frac{dy}{dx} = \frac{1}{\frac{du}{dx}} \cdot \frac{du}{dy} = \frac{1}{2\sin 2y \cos 2y} \cdot 2\cos 2y$

This simplifies to:

$\frac{dy}{dx} = \frac{1}{\sin 2y}$

Now, we can write our answer in the required form:

$\frac{dy}{dx} = \csc(2y)$

Therefore, in this case, $p = 1$ and $q = 2$ .

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 6 - 2016 - Paper 3

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 6 - 2016 - Paper 3

Find the turning point of the curve with equation $y = e^x \cos 4x$

Given $x = \sin^2 2y$, find $\frac{dy}{dx}$ as a function of $y$

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