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Figure 3 shows a sketch of part of the curve with equation $y = 1 - 2 ext{cos} \, x$, where $x$ is measured in radians - Edexcel - A-Level Maths Pure - Question 18 - 2013 - Paper 1

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Question 18

Figure-3-shows-a-sketch-of-part-of-the-curve-with-equation-$y-=-1---2-ext{cos}-\,-x$,-where-$x$-is-measured-in-radians-Edexcel-A-Level Maths Pure-Question 18-2013-Paper 1.png

Figure 3 shows a sketch of part of the curve with equation $y = 1 - 2 ext{cos} \, x$, where $x$ is measured in radians. The curve crosses the $x$-axis at the point $... show full transcript

Worked Solution & Example Answer:Figure 3 shows a sketch of part of the curve with equation $y = 1 - 2 ext{cos} \, x$, where $x$ is measured in radians - Edexcel - A-Level Maths Pure - Question 18 - 2013 - Paper 1

Step 1

Find, in terms of π, the x coordinate of the point A and the x coordinate of the point B

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Answer

To find the coordinates of points AA and BB, we need to set the equation equal to zero:

12cosx=01 - 2\text{cos} \, x = 0

From this, we obtain:

cosx=12\text{cos} \, x = \frac{1}{2}

The solutions for xx occur at:

x=π3+2nπandx=π3+2nπx = \frac{\pi}{3} + 2n\pi \quad \text{and} \quad x = -\frac{\pi}{3} + 2n\pi

Considering the principal values:

  1. AA: x=π3x = \frac{\pi}{3}
  2. BB: x=5π3x = \frac{5\pi}{3}

Step 2

Find, by integration, the exact value of the volume of the solid generated

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Answer

To find the volume of the solid generated by rotating region SS about the xx-axis, we use the formula for volume:

V=πab[f(x)]2dxV = \pi \int_{a}^{b} [f(x)]^2 \, dx

Here, f(x)=12cosxf(x) = 1 - 2\text{cos} \, x and the limits of integration are from AA to BB, which are rac{\pi}{3} and rac{5\pi}{3} respectively:

V=ππ35π3(12cosx)2dxV = \pi \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} \left(1 - 2\text{cos} \, x\right)^2 \, dx

Now, calculating the integral:

  1. Expand the square: (12cosx)2=14cosx+4cos2x\left(1 - 2\text{cos} \, x\right)^2 = 1 - 4\text{cos} \, x + 4\text{cos}^2 \, x

  2. Therefore, we get: V=ππ35π3(14cosx+41+cos(2x)2)dxV = \pi \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} \left(1 - 4\text{cos} \, x + 4\frac{1 + \text{cos}(2x)}{2}\right) \, dx

  3. Simplifying this results in: V=ππ35π3(32cosx+2cos(2x))dxV = \pi \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} \left(3 - 2\text{cos} \, x + 2\text{cos}(2x)\right) \, dx

  4. Evaluating the integral produces the final volume value.

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