Rate of Change of Triangle Area with Changing Sides and Angle
To solve the given problem, we must determine the rate of change of the area of a triangle with sides A and B and angle B between them, given how A and B are changing with time.
We are given:
- The area of a triangle is given by the formula \( A = \frac{1}{2} ab \sin{\theta} \), where \( a \) and \( b \) are the sides of the triangle and \( \theta \) is the angle between them. For our problem, we can rewrite the given area formula as \( Area = \frac{1}{2} AB \sin{B} \).
- Side \( A \) is increasing at 0.1 units/s.
- Side \( B \) is decreasing at 0.1 units/s.
- Angle \( B \) is increasing at 0.2 radians/s.
We want to find the rate of change of the triangle's area, which can be found by differentiating the area formula with respect to time \( t \). This is an application of the chain rule in calculus.
So we differentiate \( Area = \frac{1}{2} AB \sin{B} \) with respect to \( t \):
\( \frac{d(Area)}{dt} = \frac{1}{2} \left[ B \sin{B} \frac{dA}{dt} + A \sin{B} \frac{dB}{dt} + AB \cos{B} \frac{dB}{dt} \right] \)
Plugging in the rates given:
\( \frac{dA}{dt} = 0.1 \, \text{units/s} \)
\( \frac{dB}{dt} = -0.1 \, \text{units/s} \)
\( \frac{dB}{dt} = 0.2 \, \text{radians/s} \)
Substitute these into the derivative:
\( \frac{d(Area)}{dt} = \frac{1}{2} \left[ B \cdot \sin{B} \cdot 0.1 + A \cdot \sin{B} \cdot (-0.1) + AB \cdot \cos{B} \cdot 0.2 \right] \)
Unfortunately, the image doesn't provide the exact values of \( A \), \( B \), and angle \( B \) at the moment of calculation, so we cannot give a numerical answer. You would need to substitute the current values of \( A \), \( B \), and angle \( B \) into the derivative to find the rate of change at that particular moment.
