Question

A triangle has base b centimeters and height h centimeters, where the height is three times the base. both b and h are functions of time t , measured in seconds. if a represents the area of the triangle, which of the following gives the rate of change of a with respect to t ?

a. dA/dt=3b cm/sec
b. dA/dt=2b db/dt cm^2/sec
c. dA/dt=3b db/dt cm/sec
d. db/dt=3b db/dt cm^2/sec

Answer 1

The correct answer for the rate of change of the area A of a triangle with respect to time t is (c) dA/dt = 3b db/dt cm/sec, considering that the height is three times the base.

Step-by-step explanation:

The rate of change of the area A of a triangle with respect to time t can be determined by differentiating the equation for the area A = 1/2 × base × height with respect to time. Given that the height h is three times the base b, the relationship can be expressed as h = 3b. By using the product rule for differentiation and the given relationship between h and b, the rate of change of area with respect to time is dA/dt = 1/2 × (3b db/dt + b d(3b)/dt) = 1/2 × 6b db/dt = 3b db/dt cm/sec, meaning the correct answer is (c).

Answer 2

The rate of change of the area of a triangle with respect to time is given by dA/dt = 3b * db/dt cm²/sec.

Step-by-step explanation:

The rate of change of the area of a triangle with respect to time can be found using the chain rule of differentiation. The formula for the area of a triangle is A = 1/2 * base * height. In this case, the height is three times the base, so we can rewrite the formula as A = 1/2 * base * (3base). Simplifying, we have A = 3/2 * base².

Taking the derivative of both sides with respect to time, we get dA/dt = 3/2 * 2base * db/dt = 3base * db/dt.

Therefore, the rate of change of the area of the triangle with respect to time is given by dA/dt = 3b * db/dt cm²/sec. The correct option is b.