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Figure 1 shows part of the curve with equation $y = \sqrt{\tan x}$ - Edexcel - A-Level Maths Pure - Question 2 - 2007 - Paper 8

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Figure 1 shows part of the curve with equation $y = \sqrt{\tan x}$. The finite region $\mathcal{R}$, which is bounded by the curve, the x-axis and the line $x = \fra... show full transcript

Worked Solution & Example Answer:Figure 1 shows part of the curve with equation $y = \sqrt{\tan x}$ - Edexcel - A-Level Maths Pure - Question 2 - 2007 - Paper 8

Step 1

Given that $y = \sqrt{\tan x}$, complete the table with the values of $y$ corresponding to $x = 0, \frac{\pi}{16}, \frac{\pi}{8}$ and $\frac{3\pi}{16}$, giving your answers to 5 decimal places.

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Answer

xx0π16\frac{\pi}{16}π8\frac{\pi}{8}3π16\frac{3\pi}{16}
yy00.446000.643500.81742

Step 2

Use the trapezium rule with all the values of $y$ in the completed table to obtain an estimate for the area of the shaded region $R$, giving your answer to 4 decimal places.

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Answer

To apply the trapezium rule, we first calculate the area using the formula:

Area=12(b1+b2)h\text{Area} = \frac{1}{2} (b_1 + b_2) h

First, we calculate the widths:

  • h=π16h = \frac{\pi}{16}
  • b1=0.44600b_1 = 0.44600, b2=0.64350b_2 = 0.64350, b3=0.81742b_3 = 0.81742, and we sum the bases:
Area=π32(0+2(0.44600+0.64350+0.81742))\text{Area} = \frac{\pi}{32} \left(0 + 2 (0.44600 + 0.64350 + 0.81742)\right)

Calculating that gives:

  • Area = 0.4726\approx 0.4726 rounded to 4 decimal places.

Step 3

Use integration to find an exact value for the volume of the solid generated.

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Answer

To find the volume of the solid generated by rotating the region R\mathcal{R} around the x-axis, we use:

V=0π4π[tanx]2dx=π0π4tanxdxV = \int_0^{\frac{\pi}{4}} \pi [\sqrt{\tan x}]^2 dx = \pi \int_0^{\frac{\pi}{4}} \tan x \, dx

Calculating this, we have:

=π[ln(sec(π4))ln(sec(0))]=π(ln(2)0)=π2ln(2).= \pi [\ln(\sec(\frac{\pi}{4})) - \ln(\sec(0))] = \pi \left(\ln(\sqrt{2}) - 0\right) = \frac{\pi}{2} \ln(2).

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