9 mark for review a triangle has base b centimeters and height h centimeters, where the height is three…

9 mark for review 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 cm²/sec c da/dt = 3b db/dt cm/sec d da/dt = 3b db/dt cm²/sec

9 mark for review 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 cm²/sec c da/dt = 3b db/dt cm/sec d da/dt = 3b db/dt cm²/sec

Answer

Explanation:

Step1: Write the area formula

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, if $y = f(u)$ and $u = g(t)$, then $\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}$. Differentiating $A=\frac{3}{2}b^{2}$ with respect to $t$ gives $\frac{dA}{dt}=3b\frac{db}{dt}$. The units of $\frac{dA}{dt}$ are $\text{cm}^2/\text{sec}$ since $A$ is in $\text{cm}^2$ and $t$ is in $\text{sec}$.

Answer:

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