Q. 20 - 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 an equilateral triangle. 

Step-by-Step Solution

Answer

The equation that relates the area A and perimeter P of an equilateral triangle is $A=\frac{1}{12\sqrt{3}}{P}^2$.

The derivative $\frac{dA}{dt}$ and $\frac{dP}{dt}$ are related by $\frac{dA}{dt}=\frac{P}{6\sqrt{3}}\frac{dP}{dt}$.

Step 1. Formula used.

The area of an equilateral triangle is $A=\frac{\sqrt{3}}{4}{a}^2$ sq. units.

The perimeter of an equilateral triangle is $P=3a$ units.

The perimeter equation can also be written as follows.

$$\begin{gathered} P=3a \\ a=\frac{P}{3} \end{gathered}$$

Step 2. Apply the value.

Apply the value $a=\frac{P}{3}$ in $A=\frac{\sqrt{3}}{4}{a}^2$ as follows.

$$\begin{gathered} A=\frac{\sqrt{3}}{4}{a}^2 \\ A=\frac{\sqrt{3}}{4}{\left(\frac{P}{3}\right)}^2 \\ A=\frac{\sqrt{3}}{4}\left(\frac{P^2}{9}\right) \\ A=\frac{1}{12\sqrt{3}}{P}^2 \end{gathered}$$

The equation that relates the area A and the perimeter P of an equilateral triangle is $A=\frac{\sqrt{3}}{4}\left(\frac{P^2}{9}\right)$.

Step 3. Apply the differentiation.

Apply the differentiation to $A=\frac{1}{12\sqrt{3}}{P}^2$ with respect to time t as follows.

$$\begin{gathered} \frac{d}{dt}\left(A\right)=\frac{d}{dt}\left(\frac{1}{12\sqrt{3}}{P}^2\right) \\ \frac{dA}{dt}=\left(\frac{2P}{12\sqrt{3}}\right)\frac{dP}{dt} \\ \frac{dA}{dt}=\left(\frac{P}{6\sqrt{3}}\right)\frac{dP}{dt} \end{gathered}$$

The derivative $\frac{dA}{dt}$ and $\frac{dP}{dt}$ are related by $\frac{dA}{dt}=\frac{P}{6\sqrt{3}}\frac{dP}{dt}$.

Step 4. Conclusion.

The equation that relates the area A and perimeter P of an equilateral triangle is $A=\frac{1}{12\sqrt{3}}{P}^2$.

The derivative $\frac{dA}{dt}$ and $\frac{dP}{dt}$ are related by $\frac{dA}{dt}=\frac{P}{6\sqrt{3}}\frac{dP}{dt}$.