Questions: Question A triangle has a base that is decreasing at a rate of 8 cm / s with the height being held constant. What is the rate of change of the area of the triangle if the height is 2 cm? Provide your answer below: The rate of change of the area of the triangle is cm^2 / s.

Solution Steps

To find the rate of change of the area of the triangle, we can use the formula for the area of a triangle, \( A = \frac{1}{2} \times \text{base} \times \text{height} \). Given that the base is decreasing at a rate of \( 8 \, \text{cm/s} \) and the height is constant at \( 2 \, \text{cm} \), we can differentiate the area with respect to time to find the rate of change of the area.

We are given that the height \( h \) of the triangle is \( 2 \, \text{cm} \) and the base \( b \) is decreasing at a rate of \( \frac{db}{dt} = -8 \, \text{cm/s} \).

The area \( A \) of a triangle is given by the formula: \[ A = \frac{1}{2} \times b \times h \]

To find the rate of change of the area with respect to time, we differentiate \( A \) with respect to \( t \): \[ \frac{dA}{dt} = \frac{1}{2} \times h \times \frac{db}{dt} \]

Substituting the known values into the differentiated equation: \[ \frac{dA}{dt} = \frac{1}{2} \times 2 \times (-8) \]

Calculating the above expression: \[ \frac{dA}{dt} = -8 \, \text{cm}^2/\text{s} \]

Final Answer