Question - Rate of Change of Triangle Area with Changing Sides and Angle
Solution:
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.
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