Question

Let \( x \) be the length of each of the equal sides of an isosceles triangle and \( \theta \) be the angle between these sides. If \( x \) is increasing at the rate \( \frac{1}{2} \) m/hour and \( \theta \) is increasing at the rate \( \frac{\pi}{3} \) rad/hour, then the rate at which area of the triangle is increasing when \( x = 12 \) m and \( \theta = \frac{\pi}{4} \) is

• A. \( \frac{5}{3} \) \( \text{m}^2/\text{hour} \)
• B. \( \frac{\sqrt{3}}{3} \) \( \text{m}^2/\text{hour} \)
• C. \( 2 \) \( \text{m}^2/\text{hour} \)
• D. \( \sqrt{3} \) \( \frac{5}{3} \) \( \text{m}^2/\text{hour} \)

Hint:

When finding the rate of change of an area with respect to time, differentiate implicitly using the chain rule and apply the given rates of change.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to find the rate of change of the area of an isosceles triangle with respect to time, given the rates of change of its sides and the angle between them.
Step 2: Key Formula or Approach:
The area \( A \) of a triangle with two sides \( x \) and included angle \( \theta \) is:
\[ A = \frac{1}{2} x^2 \sin \theta \] Differentiating with respect to time \( t \):
\[ \frac{dA}{dt} = \frac{1}{2} \left[ 2x \frac{dx}{dt} \sin \theta + x^2 \cos \theta \frac{d\theta}{dt} \right] \]
Step 3: Detailed Explanation:
Given:
\( x = 12 \text{ m} \), \( \frac{dx}{dt} = \frac{1}{12} \text{ m/hr} \)
\( \theta = \frac{\pi}{4} \), \( \frac{d\theta}{dt} = \frac{\pi}{180} \text{ rad/hr} \)
Substitute these values into the differentiated formula:
\[ \frac{dA}{dt} = x \frac{dx}{dt} \sin \theta + \frac{1}{2} x^2 \cos \theta \frac{d\theta}{dt} \]
\[ \frac{dA}{dt} = (12) \left( \frac{1}{12} \right) \sin\left(\frac{\pi}{4}\right) + \frac{1}{2} (12^2) \cos\left(\frac{\pi}{4}\right) \left( \frac{\pi}{180} \right) \]
\[ \frac{dA}{dt} = 1 \cdot \frac{1}{\sqrt{2}} + \frac{1}{2} (144) \cdot \frac{1}{\sqrt{2}} \cdot \frac{\pi}{180} \]
\[ \frac{dA}{dt} = \frac{1}{\sqrt{2}} + \frac{72\pi}{180\sqrt{2}} = \frac{1}{\sqrt{2}} + \frac{2\pi}{5\sqrt{2}} \]
Factor out \( \frac{1}{\sqrt{2}} \):
\[ \frac{dA}{dt} = \frac{1}{\sqrt{2}} \left( 1 + \frac{2\pi}{5} \right) \]
To match the options, multiply and divide by \( \sqrt{2} \):
\[ \frac{dA}{dt} = \frac{\sqrt{2}}{2} \left( 1 + \frac{2\pi}{5} \right) = \sqrt{2} \left( \frac{1}{2} + \frac{\pi}{5} \right) \]
Step 4: Final Answer:
The rate of increase in area is \( \sqrt{2} (\frac{\pi}{5} + \frac{1}{2}) \text{m}^2/\text{hour} \).