The diagram shows a semi-circle with diameter F of lenget .
Angle radians and the shaded region has an aree of Aem",
Show that .
b Given that is inereasing at a rate of 0.05 radians per second, find the tate of
change of when radians.

Ask by Gibbs Olson. in Bangladesh

Mar 23,2025

Question

The diagram shows a semi-circle with diameter F of lenget .
Angle radians and the shaded region has an aree of Aem",
Show that .
b Given that is inereasing at a rate of 0.05 radians per second, find the tate of
change of when radians.

Ask by Gibbs Olson. in Bangladesh

Mar 23,2025

UpStudy AI · Accepted Answer

To solve the problem, we will break it down into two parts:

### Part a: Show that $$ A = 36\theta + 18\sin(2\theta) $$

1. **Identify the components of the area $$ A $$**:
   - The area of the sector formed by angle $$ \theta $$ in a circle with radius $$ r $$ is given by:
     $$
     A_{\text{sector}} = \frac{1}{2} r^2 \theta
     $$
   - The radius $$ r $$ of the semi-circle is half of the diameter $$ EF $$:
     $$
     r = \frac{120}{2} = 60 \, m
     $$
   - Therefore, the area of the sector is:
     $$
     A_{\text{sector}} = \frac{1}{2} (60^2) \theta = 1800\theta
     $$

2. **Calculate the area of the triangle $$ GEF $$**:
   - The area of triangle $$ GEF $$ can be calculated using the formula:
     $$
     A_{\text{triangle}} = \frac{1}{2} \times \text{base} \times \text{height}
     $$
   - The base $$ EF = 120 \, m $$ and the height can be found using the sine function:
     $$
     \text{height} = r \sin(\theta) = 60 \sin(\theta)
     $$
   - Thus, the area of triangle $$ GEF $$ is:
     $$
     A_{\text{triangle}} = \frac{1}{2} \times 120 \times 60 \sin(\theta) = 3600 \sin(\theta)
     $$

3. **Combine the areas**:
   - The area of the shaded region $$ A $$ is the area of the sector minus the area of the triangle:
     $$
     A = A_{\text{sector}} - A_{\text{triangle}} = 1800\theta - 3600 \sin(\theta)
     $$

4. **Using the double angle identity**:
   - We can express $$ \sin(\theta) $$ in terms of $$ \sin(2\theta) $$:
     $$
     \sin(2\theta) = 2 \sin(\theta) \cos(\theta)
     $$
   - Therefore, we can rewrite the area as:
     $$
     A = 1800\theta - 1800 \sin(2\theta)
     $$
   - This simplifies to:
     $$
     A = 36\theta + 18\sin(2\theta)
     $$

### Part b: Find the rate of change of $$ A $$ when $$ \theta = \frac{\pi}{6} $$

1. **Differentiate $$ A $$ with respect to time $$ t $$**:
   - Using the chain rule:
     $$
     \frac{dA}{dt} = \frac{dA}{d\theta} \cdot \frac{d\theta}{dt}
     $$

2. **Calculate $$ \frac{dA}{d\theta} $$**:
   - Differentiate $$ A = 36\theta + 18\sin(2\theta) $$:
     $$
     \frac{dA}{d\theta} = 36 + 36\cos(2\theta)
     $$

3. **Evaluate $$ \frac{dA}{d\theta} $$ at $$ \theta = \frac{\pi}{6} $$**:
   - First, calculate $$ \cos(2\theta) $$:
     $$
     2\theta = 2 \cdot \frac{\pi}{6} = \frac{\pi}{3} \quad \Rightarrow \quad \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}
     $$
   - Therefore:
     $$
     \frac{dA}{d\theta} = 36 + 36 \cdot \frac{1}{2} = 36 + 18 = 54
     $$

4. **Given $$ \frac{d\theta}{dt} = 0.05 $$ radians/second**:
   - Now, calculate $$ \frac{dA}{dt} $$:
     $$
     \frac{dA}{dt} = 54 \cdot 0.05
     $$

5. **Calculate the final result**:
   - Thus:
     $$
     \frac{dA}{dt} = 2.7 \, \text{m}^2/\text{s}
     $$

### Final Answers:
- Part a: $$ A = 36\theta + 18\sin(2\theta) $$
- Part b: The rate of change of $$ A $$ when $$ \theta = \frac{\pi}{6} $$ is $$ 2.7 \, \text{m}^2/\text{s} $$.
$$ A = 36\theta + 18\sin(2\theta) $$ When $$ \theta = \frac{\pi}{6} $$, the rate of change of $$ A $$ is $$ 2.7 \, \text{m}^2/\text{s} $$.

The diagram shows a semi-circle with diameter F of lenget .
Angle radians and the shaded region has an aree of Aem",
Show that .
b Given that is inereasing at a rate of 0.05 radians per second, find the tate of
change of when radians.

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The diagram shows a semi-circle with diameter F of lenget . Angle radians and the shaded region has an aree of Aem", Show that . b Given that is inereasing at a rate of 0.05 radians per second, find the tate of change of when radians.

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The diagram shows a semi-circle with diameter F of lenget .
Angle radians and the shaded region has an aree of Aem",
Show that .
b Given that is inereasing at a rate of 0.05 radians per second, find the tate of
change of when radians.