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) (\frac{da}{dt}=3b cm/sec) (b) (\frac{da}{dt}=2b\frac{db}{dt} cm^{2}/sec) (c) (\frac{da}{dt}=3b\frac{db}{dt} cm/sec) (d) (\frac{da}{dt}=3b\frac{db}{dt} cm^{2}/sec)

Explanation:

Step1: Write the area formula for a triangle

The area formula of a triangle is $A=\frac{1}{2}bh$. Given that $h = 3b$, then $A=\frac{1}{2}b\times(3b)=\frac{3}{2}b^{2}$.

Step2: Differentiate with respect to $t$

Using the chain - rule $\frac{dA}{dt}=\frac{dA}{db}\cdot\frac{db}{dt}$. Differentiating $A=\frac{3}{2}b^{2}$ with respect to $b$ gives $\frac{dA}{db}=3b$. Then $\frac{dA}{dt}=3b\frac{db}{dt}$. The units of $A$ are $\text{cm}^2$ and of $t$ is $\text{sec}$, so the units of $\frac{dA}{dt}$ are $\text{cm}^2/\text{sec}$.

Answer:

D. $\frac{dA}{dt}=3b\frac{db}{dt}\text{ cm}^2/\text{sec}$