Chapter 3: Q. 25 (page 299)

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.

Short Answer

Expert verified

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

Implicit differentiation: $\frac{d A}{d t} = \frac{c}{2} \frac{d c}{d t}$

Step by step solution

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,

${(b a s e)}^{2} + {(h e i g h t)}^{2} = {(h y o t e n u s e)}^{2}, where, b a s e = h e i g h t Let, b a s e = h e i g h t = b \Rightarrow b^{2} + b^{2} = c^{2} \Rightarrow 2 b^{2} = c^{2} \Rightarrow b^{2} = \frac{c^{2}}{2} \Rightarrow b = \frac{c}{\sqrt{2}} The area of the isosceles right angled triangle is given by, A = \frac{(b a s e) (h e i g h t)}{2} \Rightarrow A = \frac{(\frac{c}{\sqrt{2}}) (\frac{c}{\sqrt{2}})}{2} \Rightarrow A = \frac{c^{2}}{4}$

Step 3. Implicit differentiation with respect to time

From the above step,

$A = \frac{c^{2}}{4} Differentiating on both sides, we get, \frac{d A}{d t} = \frac{2 c}{4} \frac{d c}{d t} \Rightarrow \frac{d A}{d t} = \frac{c}{2} \frac{d c}{d t}$