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 hypotenuse c of a triangle that is similar to a right triangle with legs of lengths 3 and 4 units and hypotenuse of length 5 units.

Short answer

Equation relating the quantities:$A=\frac{6}{25}{c}^2$

Implicit differentiation:$\frac{dA}{dt}=\frac{12c}{25}\frac{dc}{dt}$

Step-by-step solution

Step 1. Given information  

Area of right angled triangle$=A$

Hypotenuse of right angled triangle role="math" localid="1648696308561" $=c=5x$

Legs of the right angled triangle are:$3x\mathrm{and}4x$

Step 2. Equation relating the quantities 

Area of the triangle is given by,

$$\begin{gathered} A=\frac{\left(base\right)\left(height\right)}{2} \\ \Rightarrow A=\frac{\left(3x\right)\left(4x\right)}{2} \\ \Rightarrow A=\frac{12x^2}{2} \\ \Rightarrow A=6x^2 \\ \mathrm{we}\mathrm{have},c=5x \\ \mathrm{Therefore},A=6{\left(\frac{c}{5}\right)}^2 \\ \Rightarrow A=\frac{6}{25}{c}^2 \end{gathered}$$

Step 3. Implicit differentiation with respect to time

From the above step,

$$\begin{gathered} A=\frac{6}{25}{c}^2 \\ \mathrm{Differentiating}\mathrm{on}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}, \\ \frac{dA}{dt}=\frac{6}{25}\left(2c\right)\frac{dc}{dt} \\ \Rightarrow \frac{dA}{dt}=\frac{12c}{25}\frac{dc}{dt} \end{gathered}$$