Question

question: the base of a triangle is shrinking at a rate of 9 cm/s and the height of the triangle is increasing at a rate of 7 cm/s. find the rate at which the area of the triangle changes when the height is 13 cm and the base is 4 cm. the rate of change of the area of the triangle is cm²/s. provide your answer below. feedback more instruction show steps submit

Explanation:

Step1: Recall the area formula for a 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 $(uv)^\prime = u^\prime v+uv^\prime$, where $u = \frac{1}{2}b$ and $v = h$. We get $\frac{dA}{dt}=\frac{1}{2}(b\frac{dh}{dt}+h\frac{db}{dt})$.

Step3: Substitute the given values

We are given that $\frac{db}{dt}=- 9$ cm/s (negative because the base is shrinking), $\frac{dh}{dt}=7$ cm/s, $b = 4$ cm and $h = 13$ cm.
Substitute these values into the formula for $\frac{dA}{dt}$:
\[

$$\begin{align*} \frac{dA}{dt}&=\frac{1}{2}(4\times7 + 13\times(-9))\\ &=\frac{1}{2}(28-117)\\ &=\frac{1}{2}(-89)\\ &=-44.5 \end{align*}$$

\]

Answer:

$-44.5$ cm²/s