Question

The base of a triangle is shrinking at a rate of 5 cm/s and the height of the triangle is increasing at a rate of 9 cm/s. Find the rate at which the area of the triangle changes when the height is 12 cm and the base is 4 cm.

Answer 1

The rate at which the area of the triangle changes when the height is 12 cm and the base is 4 cm is -12 cm²/s.

Step-by-step explanation:

To find the rate at which the area of the triangle changes, we can use the formula for the area of a triangle, which is ½ * base * height. Let's denote the base of the triangle as b and the height as h. The given rates are db/dt = -5 cm/s (the base is shrinking) and dh/dt = 9 cm/s (the height is increasing).

At the specific moment when the height is 12 cm and the base is 4 cm, we can substitute these values into our area formula to find the answer. So, area = ½ * (base) * (height) = ½ * (4 cm) * (12 cm) = 24 cm².

To find the rate at which the area changes, we can differentiate the area formula concerning time. dA/dt = (½ * h) * (db/dt) + (½ * b) * (dh/dt). Now we substitute the given values into this formula: dA/dt = (½ * 12 cm) * (-5 cm/s) + (½ * 4 cm) * (9 cm/s) = -30 cm²/s + 18 cm²/s = -12 cm²/s.