Q. 25 - 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 hypotenuse c of an isosceles right triangle.  

Step-by-Step Solution

Answer

Equation relating the quantities: $A=\frac{c^2}{4}$

Implicit differentiation: $\frac{dA}{dt}=\frac{c}{2}\frac{dc}{dt}$

Step 1. Given information

Area of isosceles right angled triangle $=A$

Hypotenuse of isosceles right angled triangle $=c$

Step 2. Equation relating the quantities

For a isosceles right angled triangle,

$$\begin{gathered} {\left(base\right)}^2+{\left(height\right)}^2={\left(hyotenuse\right)}^2, \mathrm{where}, base = height \\ \mathrm{Let}, base = height =b \\ \Rightarrow b^2+b^2=c^2 \\ \Rightarrow 2b^2=c^2 \\ \Rightarrow b^2=\frac{c^2}{2} \\ \Rightarrow b=\frac{c}{\sqrt{2}} \\ \mathrm{The} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{isosceles} \mathrm{right} \mathrm{angled} \mathrm{triangle} \mathrm{is} \mathrm{given} \mathrm{by}, \\ A=\frac{\left(base\right)\left(height\right)}{2} \\ \Rightarrow A=\frac{\left(\frac{c}{\sqrt{2}}\right)\left(\frac{c}{\sqrt{2}}\right)}{2} \\ \Rightarrow A=\frac{c^2}{4} \end{gathered}$$

Step 3. Implicit differentiation with respect to time

From the above step,

$$\begin{gathered} A=\frac{c^2}{4} \\ \mathrm{Differentiating} \mathrm{on} \mathrm{both} \mathrm{sides}, \mathrm{we} \mathrm{get}, \\ \frac{dA}{dt}=\frac{2c}{4}\frac{dc}{dt} \\ \Rightarrow \frac{dA}{dt}=\frac{c}{2}\frac{dc}{dt} \end{gathered}$$