Q. 19 - Calculus

Question

In Exercises 19–26, write down an equation that relates the two quantities described. Then use implicit differentiation to obtain a relationship between the rates at which the quantities change over time.

The area A and perimeter P of a square. 

Step-by-Step Solution

Answer

The equation that relates the area A and perimeter P of a square is $A=\frac{P^2}{16}$.

The derivative $\frac{dA}{dt}$ and $\frac{dP}{dt}$ are related by $\frac{dA}{dt}=\frac{P}{8}\left(\frac{dP}{dt}\right)$.

Step 1. Formula used.

The area of a square is $A=a^2$ sq. units.

The perimeter of a square is $P=4a$ units.

The perimeter can be written as follows.

$$\begin{gathered} P=4a \\ a=\frac{P}{4} \end{gathered}$$

Step 2. Apply the value.

The area A and perimeter P of the square are related by applying $a=\frac{P}{4}$ in $A=a^2$ as follows.

$$\begin{gathered} A=a^2 \\ A={\left(\frac{P}{4}\right)}^2 \\ A=\frac{P^2}{16} \end{gathered}$$

The equation that relates the area A and perimeter P of the square is $A=\frac{P^2}{16}$.

Step 3. Apply the differentiation.

Apply the differentiation to $A=\frac{P^2}{16}$ with respect to time t as follows.

The derivatives $\frac{dA}{dt}$ and $\frac{dP}{dt}$ are related by $\frac{dA}{dt}=\frac{P}{8}\left(\frac{dP}{dt}\right)$.

Step 4. Conclusion.

The equation that relates the area A and perimeter P of the square is $A=\frac{P^2}{16}$.

The derivative $\frac{dA}{dt}$ and $\frac{dP}{dt}$ are related by $\frac{dA}{dt}=\frac{P}{8}\left(\frac{dP}{dt}\right)$.