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. provide your answer below: the rate of change of the area of the triangle is □ cm²/s.

Explanation:

Step1: Recall area formula for triangle

The area formula of a triangle is $A=\frac{1}{2}bh$, where $b$ is the base and $h$ is the height.

Step2: Differentiate with respect to time $t$

Using the product - rule $\frac{d(uv)}{dt}=u\frac{dv}{dt}+v\frac{du}{dt}$, we have $\frac{dA}{dt}=\frac{1}{2}(b\frac{dh}{dt}+h\frac{db}{dt})$.

Step3: Substitute given values

We are given that $\frac{db}{dt}=- 5$ cm/s (negative because the base is shrinking), $\frac{dh}{dt}=9$ cm/s, $b = 4$ cm and $h = 12$ cm.
Substitute these values into the formula for $\frac{dA}{dt}$:
\[

$$\begin{align*} \frac{dA}{dt}&=\frac{1}{2}(4\times9 + 12\times(-5))\\ &=\frac{1}{2}(36-60)\\ &=\frac{1}{2}\times(-24)\\ &=-12 \end{align*}$$

\]

Answer:

$-12$ cm$^{2}$/s