Final answer:
The rate at which the area of the triangle changes when the base is 10 cm and the height is 14 cm is -46 cm²/s, meaning the area is decreasing at this rate.
Step-by-step explanation:
The student is asking about related rates, a concept in calculus that deals with how the rate of change of one quantity is related to the rate of change of another quantity. In this problem, we're given that the base b of a triangle is shrinking at a rate of 8 cm/s (db/dt = -8 cm/s since it's decreasing) and the height h is increasing at a rate of 2 cm/s (dh/dt = 2 cm/s). We're asked to find the rate at which the area A of the triangle changes when h is 14 cm and b is 10 cm.
The area of a triangle is given by the formula A = 1/2 × b × h. To find the rate of change of the area, dA/dt, we'll differentiate both sides of the equation with respect to time t:
dA/dt = 1/2 × (b × dh/dt + h × db/dt)
Plugging in the given values:
dA/dt = 1/2 × (10 cm × 2 cm/s + 14 cm × -8 cm/s)
dA/dt = 1/2 × (20 cm^2/s - 112 cm^2/s)
dA/dt = 1/2 × (-92 cm^2/s)
dA/dt = -46 cm^2/s
So, the area is decreasing at a rate of 46 cm2/s when the base is 10 cm and the height is 14 cm.