A curve has equation $y = 2x \, ext{cos} \, 3x + (3x^2 - 4) \, ext{sin} \, 3x.$ 8 (a) Find $\frac{dy}{dx},$ giving your answer in the form $(mx^2 + n) \text{cos} \, 3x,$ where $m$ and $n$ are integers. 8 (b) Show that the x-coordinates of the points of inflection of the curve satisfy the equation $\text{cot} \, 3x = \frac{9x^2 - 10}{6x}.$

A-Level AQA Maths Pure Trigonometric Functions: A curve has equation $y = 2x \

A curve has equation $y = 2x \, ext{cos} \, 3x + (3x^2 - 4) \, ext{sin} \, 3x.$ 8 (a) Find $\frac{dy}{dx},$ giving your answer in the form $(mx^2 + n) \text{cos} \, 3x,$ where $m$ and $n$ are integers - AQA - A-Level Maths Pure - Question 8 - 2017 - Paper 2

Question 8

$A-curve-has-equation---$y-=-2x-\,--ext{cos}-\,-3x-+-(3x^2---4)-\,--ext{sin}-\,-3x.$----8-(a)-Find-$\frac{dy}{dx},$-giving-your-answer-in-the-form-$(mx^2-+-n)-\text{cos}-\,-3x,$-where-$m$-and-$n$-are-integers-AQA-A-Level Maths Pure-Question 8-2017-Paper 2.png$

A curve has equation $y = 2x \, ext{cos} \, 3x + (3x^2 - 4) \, ext{sin} \, 3x.$ 8 (a) Find $\frac{dy}{dx},$ giving your answer in the form $(mx^2 + n) \text{c... show full transcript

Worked Solution & Example Answer:A curve has equation $y = 2x \, ext{cos} \, 3x + (3x^2 - 4) \, ext{sin} \, 3x.$ 8 (a) Find $\frac{dy}{dx},$ giving your answer in the form $(mx^2 + n) \text{cos} \, 3x,$ where $m$ and $n$ are integers - AQA - A-Level Maths Pure - Question 8 - 2017 - Paper 2

Step 1

Find $\frac{dy}{dx},$ giving your answer in the form $(mx^2 + n) \text{cos} \, 3x$

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Answer

To find the derivative $\frac{dy}{dx},$ we will use the product rule. The equation of the curve is:

$y = 2x \cos 3x + (3x^2 - 4) \sin 3x.$

We differentiate each term separately:

For the first term, $2x \cos 3x$ , we apply the product rule: $\frac{d}{dx}(2x \cos 3x) = 2 \cos 3x - 2x(3 \sin 3x) = 2 \cos 3x - 6x \sin 3x.$
For the second term, $(3x^2 - 4) \sin 3x$ , we again apply the product rule: $\frac{d}{dx}((3x^2 - 4) \sin 3x) = (6x) \sin 3x + (3x^2 - 4)(3 \cos 3x) = 6x \sin 3x + 3(3x^2 - 4) \cos 3x.$

Next, we combine these results:

$\frac{dy}{dx} = (2 \cos 3x - 6x \sin 3x) + (6x \sin 3x + 3(3x^2 - 4) \cos 3x).$

The terms $-6x \sin 3x$ and $6x \sin 3x$ cancel out, yielding:

$\frac{dy}{dx} = (2 + 3(3x^2 - 4)) \cos 3x = (9x^2 - 10) \cos 3x.$

Thus, we have expressed $\frac{dy}{dx}$ in the desired form: where $m = 9$ and $n = -10.$

Step 2

Show that the x-coordinates of the points of inflection of the curve satisfy the equation $\text{cot} \, 3x = \frac{9x^2 - 10}{6x}$

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Answer

To find the x-coordinates of the points of inflection, we need to set the second derivative $\frac{d^2y}{dx^2}$ to zero.

First, we differentiate the first derivative $\frac{dy}{dx}$ :

$\frac{dy}{dx} = (9x^2 - 10) \cos 3x.$

Using the product rule again, we have:

$\frac{d^2y}{dx^2} = \frac{d}{dx}((9x^2 - 10) \cos 3x) = (18x) \cos 3x + (9x^2 - 10)(-3 \sin 3x).$

Setting this equal to zero for points of inflection:

$0 = (18x \cos 3x - 3(9x^2 - 10) \sin 3x).$

Rearranging gives:

$18x \cos 3x = 3(9x^2 - 10) \sin 3x.$

Dividing both sides by $3x \cos 3x$ (assuming $x \neq 0$ ), we get:

$6 = \frac{(9x^2 - 10) \sin 3x}{x \cos 3x}.$

Taking cotangent of both sides gives:

$\text{cot} \, 3x = \frac{9x^2 - 10}{6x},$

thus proving the relationship.

A curve has equation $y = 2x \, ext{cos} \, 3x + (3x^2 - 4) \, ext{sin} \, 3x.$ 8 (a) Find $\frac{dy}{dx},$ giving your answer in the form $(mx^2 + n) \text{cos} \, 3x,$ where $m$ and $n$ are integers - AQA - A-Level Maths Pure - Question 8 - 2017 - Paper 2

Worked Solution & Example Answer:A curve has equation $y = 2x \, ext{cos} \, 3x + (3x^2 - 4) \, ext{sin} \, 3x.$ 8 (a) Find $\frac{dy}{dx},$ giving your answer in the form $(mx^2 + n) \text{cos} \, 3x,$ where $m$ and $n$ are integers - AQA - A-Level Maths Pure - Question 8 - 2017 - Paper 2

Find $\frac{dy}{dx},$ giving your answer in the form $(mx^2 + n) \text{cos} \, 3x$

Show that the x-coordinates of the points of inflection of the curve satisfy the equation $\text{cot} \, 3x = \frac{9x^2 - 10}{6x}$

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