Chapter 2: Q. 86 (page 211)

$If you drop a pebble into a large lake, you will cause a circle of ripples to expand outward . The area A = A (t) and radius r = r (t) are clearly functions of t (they change over time) and are related by the formula A = {πr}^{2} . (a) If r is measured in inches and t is measured in seconds, what are the units of \frac{d A}{d t} ? What are the units of \frac{d A}{d r} ? (b) Find \frac{d A}{d r} and explain the practical meaning of {\frac{d A}{d r}}_{r = 2} . (c) Find \frac{d A}{d t} and explain the practical meaning of {\frac{d A}{d t}}_{r = 2} .$

Short Answer

Expert verified

$(a) . Unit of \frac{d A}{d t} = inche per seconds, unit of \frac{d A}{d r} = inch . (b) . The value of \frac{d A}{d r} = 2 πr . (c) . The value of \frac{d A}{d t} = 2 πr \frac{d r}{d t}$

Step by step solution

Step 1. Given Information

$A = {πr}^{2}$

Step 2. Solution (a) : The units of dAdt and dAdr

$Consider the equation : A = {πr}^{2} Differentiate above equation : \frac{d A}{d t} = \frac{d}{d t} ({πr}^{2}) = π \frac{d}{d t} (r^{2}) = 2 πr \frac{d r}{d t} inch per seconds . . . . (1) Consider the equation : A = {πr}^{2} Differentiate above equation \frac{d A}{d r} = \frac{d}{d r} ({πr}^{2}) = π \frac{d}{dr} (r^{2}) = 2 πr inch . . . . . . (2)$

Step 3. Solution (b) : value of dAdr.

$Consider the equation : A = {πr}^{2} Differentiate above equation \frac{d A}{d r} = \frac{d}{d r} ({πr}^{2}) = π \frac{d}{dr} (r^{2}) = 2 πr inch The practical meaning is the rate of change of area of a circle with radius 2 .$

Step 4. Solution (c) : value of dAdt

$Consider the equation : A = {πr}^{2} Differentiate above equation : \frac{d A}{d t} = \frac{d}{d t} ({πr}^{2}) = π \frac{d}{d t} (r^{2}) = 2 πr \frac{d r}{d t} inch per seconds The practical meaning is the rate of change of area of circle with respect to time radius 2 .$