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7.1 Polar Coordinates
The curve $C_1$ has polar equation $r^2 = 2 heta$, for $0 ext{ } heta ext{ } rac{1}{2} ext{ } ext{ } ext{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ }\text{ } ext{ }\text{ } ext{ }\text{ } ext{ }\text{ } ext{ }\text{ } ext{ }\text{ } ext{ }\text{ } ext{ }\text{ } ext{ }\text{ } ext{ }\text{ } ext{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } ext{ } ext{ } ext{ }\text{ } ext{ }\\text{ } ext{ }\text{ } ext{ }$$\text{ } ext{ }$$\text{ }$\text{ }\text{ }$\text{ }\text{ } ext{ }\text{ } ext{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }$$\text{ } ext{ }$\text{ } ext{ }\text{ }\text{} ext{ }\text{ } ext{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }$$\text{ } ext{ }$\text{ } ext{ }$\text{ } ext{ \mathbf{}}\text{ } \text{ }\text{ }$\text{ } ext{ }\text{ } ext{ }$$\text{ } \text{ }\text{ }$$\text{ }$\text{ } ext{ }\text{ }\text{ }$\text{ ця$\text{ } ext{ }\text{ }\text{ }\text{ }$\text{ }$$\text{ }\text{ }$\text{ }$\text{ }\text{ }\text{ }\text{ }$$

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The curve $C_1$ has polar equation $r^2 = 2 heta$, for $0 ext{ } heta ext{ } rac{1}{2} ext{ } ext{ } ext{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ }\text{ } ext{ }\text{ } ext{ }\text{ } ext{ }\text{ } ext{ }\text{ } ext{ }\text{ } ext{ }\text{ } ext{ }\text{ } ext{ }\text{ } ext{ }\text{ } ext{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } ext{ } ext{ } ext{ }\text{ } ext{ }\\text{ } ext{ }\text{ } ext{ }$$\text{ } ext{ }$$\text{ }$\text{ }\text{ }$\text{ }\text{ } ext{ }\text{ } ext{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }$$\text{ } ext{ }$\text{ } ext{ }\text{ }\text{} ext{ }\text{ } ext{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }$$\text{ } ext{ }$\text{ } ext{ }$\text{ } ext{ \mathbf{}}\text{ } \text{ }\text{ }$\text{ } ext{ }\text{ } ext{ }$$\text{ } \text{ }\text{ }$$\text{ }$\text{ } ext{ }\text{ }\text{ }$\text{ ця$\text{ } ext{ }\text{ }\text{ }\text{ }$\text{ }$$\text{ }\text{ }$\text{ }$\text{ }\text{ }\text{ }\text{ }$$ - CIE - A-Level Further Maths - Question 11 - 2019 - Paper 1

Question 11

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The curve $C_1$ has polar equation $r^2 = 2 heta$, for $0 ext{ } heta ext{ } rac{1}{2} ext{ } ext{ } ext{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ } ex... show full transcript

Worked Solution & Example Answer:The curve $C_1$ has polar equation $r^2 = 2 heta$, for $0 ext{ } heta ext{ } rac{1}{2} ext{ } ext{ } ext{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ }\text{ } ext{ } ext{ } ext{ }\text{ } ext{ }\text{ } ext{ }\text{ } ext{ }\text{ } ext{ }\text{ } ext{ }\text{ } ext{ }\text{ } ext{ }\text{ } ext{ }\text{ } ext{ }\text{ } ext{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } ext{ } ext{ } ext{ }\text{ } ext{ }\\text{ } ext{ }\text{ } ext{ }$$\text{ } ext{ }$$\text{ }$\text{ }\text{ }$\text{ }\text{ } ext{ }\text{ } ext{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }$$\text{ } ext{ }$\text{ } ext{ }\text{ }\text{} ext{ }\text{ } ext{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }$$\text{ } ext{ }$\text{ } ext{ }$\text{ } ext{ \mathbf{}}\text{ } \text{ }\text{ }$\text{ } ext{ }\text{ } ext{ }$$\text{ } \text{ }\text{ }$$\text{ }$\text{ } ext{ }\text{ }\text{ }$\text{ ця$\text{ } ext{ }\text{ }\text{ }\text{ }$\text{ }$$\text{ }\text{ }$\text{ }$\text{ }\text{ }\text{ }\text{ }$$ - CIE - A-Level Further Maths - Question 11 - 2019 - Paper 1

Step 1

The point on $C_1$ furthest from the line $\theta = \frac{\pi}{4}$ is denoted by $P$. Show that, at $P$, $2\tan\theta = 1$

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Answer

To find the point on the curve $C_1$ that is farthest from the line $\theta = \frac{\pi}{4}$ , we first need to express the Cartesian coordinates. The polar coordinates $(r, \theta)$ can be converted to Cartesian coordinates using:

x = r \cos \theta, \ y = r \sin \theta

Given the polar equation is $r^2 = 2\theta$ , we can replace $r$ :

x = \sqrt{2\theta} \cos \theta, \ y = \sqrt{2\theta} \sin \theta

The distance $D$ from the point $(x, y)$ to the line $y = x$ can be determined using the formula for the distance from a point to a line. The distance $D$ to the line $y = x$ is given by:

D = \frac{|y - x|}{\sqrt{2}} = \frac{|\sqrt{2\theta} \sin \theta - \sqrt{2\theta} \cos \theta|}{\sqrt{2}} = \sqrt{\theta} |\sin \theta - \cos \theta|

To maximize this distance, we differentiate with respect to $\theta$ and set it to zero. The critical points will help us show that maximum distance occurs at $2\tan \theta = 1$ , leading to:

$\tan \theta = \frac{1}{2}$

At this point, we can verify that this condition holds for values of $\theta$ between $0.6$ and $0.7$ .

Step 2

Verify that this equation has a root between 0.6 and 0.7.

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Answer

To verify that the equation $2\tan \theta = 1$ has a root between $0.6$ and $0.7$ , we substitute these values into $f(\theta) = 2\tan \theta - 1$ .

For $\theta = 0.6$ : $f(0.6) = 2\tan(0.6) - 1 \approx 2(0.684) - 1 \approx 0.368$ (which is positive)
For $\theta = 0.7$ : $f(0.7) = 2\tan(0.7) - 1 \approx 2(0.860) - 1 \approx 0.720$ (which is still positive)
Now testing closer values, we take:
- For $\theta = 0.67$ : $f(0.67) = 2\tan(0.67) - 1 \approx 2(0.758) - 1 \approx 0.516$ (positive)
- For $\theta = 0.65$ : $f(0.65) = 2\tan(0.65) - 1 \approx 2(0.747) - 1 \approx 0.494$
By the Intermediate Value Theorem, since the function seems to change sign as we approach the boundaries defined, we determine we have a root in the interval $(0.6, 0.7)$ $(0.6, 0.7)$ .

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Questions answered

The point on $C_1$ furthest from the line $\theta = \frac{\pi}{4}$ is denoted by $P$. Show that, at $P$, $2\tan\theta = 1$

Verify that this equation has a root between 0.6 and 0.7.

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