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A curve C has equation $y = 3 ext{sin}2x + 4 ext{cos}2x, - ext{π} < x < ext{π}$ - Edexcel - A-Level Maths Pure - Question 7 - 2008 - Paper 6

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A-curve-C-has-equation--$y-=-3-ext{sin}2x-+-4-ext{cos}2x,---ext{π}-<-x-<--ext{π}$-Edexcel-A-Level Maths Pure-Question 7-2008-Paper 6.png

A curve C has equation $y = 3 ext{sin}2x + 4 ext{cos}2x, - ext{π} < x < ext{π}$. The point A(0, 4) lies on C. (a) Find an equation of the normal to the curve C a... show full transcript

Worked Solution & Example Answer:A curve C has equation $y = 3 ext{sin}2x + 4 ext{cos}2x, - ext{π} < x < ext{π}$ - Edexcel - A-Level Maths Pure - Question 7 - 2008 - Paper 6

Step 1

Find an equation of the normal to the curve C at A.

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Answer

To find the normal at point A(0, 4), we first need to compute the derivative of the curve:

dydx=6cos(2x)8sin(2x)\frac{dy}{dx} = 6\cos(2x) - 8\sin(2x)

Evaluating this at x = 0 gives:

dydxx=0=6cos(0)8sin(0)=6\frac{dy}{dx} \bigg|_{x=0} = 6\cos(0) - 8\sin(0) = 6

The slope of the normal line is the negative reciprocal of the slope of the tangent line:

mn=16m_n = -\frac{1}{6}

Using the point-slope form of the line, the equation of the normal line is:

y4=16(x0)y - 4 = -\frac{1}{6}(x - 0)

Simplifying gives:

y=16x+4y = -\frac{1}{6}x + 4

Step 2

Express y in the form Rsin(2x + α), where R > 0 and 0 < α < π/2.

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Answer

To express y=3sin(2x)+4cos(2x)y = 3\text{sin}(2x) + 4\text{cos}(2x) in the desired form, we need to find R and α:

We compute:

R=32+42=9+16=25=5R = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5

Next, we find α:

tan(α)=43\tan(\alpha) = \frac{4}{3}

This gives:

α=arctan(43)0.927\alpha = \arctan\left(\frac{4}{3}\right) \approx 0.927 (to 3 significant figures).

Step 3

Find the coordinates of the points of intersection of the curve C with the x-axis.

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Answer

The points of intersection with the x-axis occur where y = 0:

Setting the equation to zero gives:

3sin(2x)+4cos(2x)=03\sin(2x) + 4\cos(2x) = 0

This can be rearranged as:

sin(2x)=43cos(2x)\sin(2x) = -\frac{4}{3}\cos(2x)

Taking the sine of both sides yields:

tan(2x)=43\tan(2x) = -\frac{4}{3}

Finding values of 2x using:

2x=arctan(43)+nπ2x = \arctan\left(-\frac{4}{3}\right) + n\pi

Calculating the values yields:

For 2x:

2x2.03,0.46,3.11,2.682x \approx -2.03, 0.46, 3.11, 2.68

Thus, the x-coordinates are:

x1.015,0.23,1.555,1.34x \approx -1.015, 0.23, 1.555, 1.34 (to 2 decimal places).

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