Question

this question has several parts that must be completed sequentially. if you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. tutorial exercise the length of a rectangle is increasing at a rate of 4 cm/s and its width is increasing at a rate of 3 cm/s. when the length is 14 cm and the width is 4 cm, how fast is the area of the rectangle increasing (in cm²/s)? step 1 let l and w represent the length and width of the rectangle in cm, and let a represent the area of the rectangle in cm². writing an equation for a in terms of l and w gives the following result. a =

Explanation:

Step1: Recall area formula

The area of a rectangle is given by the product of its length and width. So, $A = lw$.

Step2: Differentiate with respect to time

Using the product - rule $\frac{d(uv)}{dt}=u\frac{dv}{dt}+v\frac{du}{dt}$, where $u = l$ and $v = w$. We have $\frac{dA}{dt}=l\frac{dw}{dt}+w\frac{dl}{dt}$.

Step3: Identify given values

We are given that $\frac{dl}{dt}=4$ cm/s, $\frac{dw}{dt}=3$ cm/s, $l = 14$ cm, and $w = 4$ cm.

Step4: Substitute values

Substitute the values into the formula for $\frac{dA}{dt}$: $\frac{dA}{dt}=14\times3 + 4\times4$.

Step5: Calculate

$\frac{dA}{dt}=42+16=58$ cm²/s.

Answer:

58