CONNECTED RATES OF CHANGE

Example , Exercises , Answers

DEFINITION OF 'RATE OF CHANGE' Rate of change is the speed at which a variable changes over a specific period of time. It's often used when speaking about momentum, and it can generally be expressed as a ratio between a change in one variable relative to a corresponding change in another. Graphically, the rate of change is represented by the slope of a line. To solve problems regarding the rate of change, we use the chain rule. EXAMPLE If a given function y = f(x) has a rate of change with respect to r; we can also find the rate of change of x with respect to r. * The volume, Vcm3, of a sphere of radius r cm is given by V = 4 3 πr3. V is increasing at the rate of 1.5cm3 per second. Find the rate of increase of r with respect to time when r = 2 ANSWER Given: dV dt = 1.5cm3 (rate of change of V) and the volume V = 4 3 πr3. Searched: dr dt Derivating V with respect to r gives us: dV dr = 4πr2 We know that: dV dr = dV dt · dt dr ⇒ 4πr2 = 1.5 · dt dr ⇒ 4πr2 1.5 = dt dr ;   Multiply both fractions by (−1) ⇒ 1.5 4πr2 = dr dt , with r = 2 ⇒ dr dt = 1.5 16π = 0.0298 cm/s Therefore, The radius increases at the rate of 0.0298 cm per second. EXERCISES The equation of a curve is y = x² − 5x. A point P is moving along the curve so that the x-coordinate is increasing at the constant rate of 0.2 units per second. Find the rate at which the y-coordinate is increasing when x = 4 Answer i. The equation of a curve is y = 4 − 1 x A point P is moving along the curve so that the y-coordinate is increasing at the constant rate of 0.01 units per second. Find the rate at which the x-coordinate is increasing when x = 1 Answer ii. The equation of a curve is y = 1 2 − x A point P is moving along the curve so that the y-coordinate is increasing at the constant rate of 0.5 units per second. Find the rate at which the x-coordinate is increasing when x = 1 Answer iii. The volume, Vcm3, of a cube of the edge xcm increasing at the constant rate of 3cm3 per second. Find the rate at which x is increasing when x = 10 The volume, Vcm3, of a sphere of radius xcm is given by V = 4 3 πx3. The volume is increasing at the constant rate of 0.2 cm3 per second. Find the rate of increase of the radius when the radius is 5cm A rectangular water tank (see below) is being filled at the constant rate of 20 l sec The base of the tank has dimensions w = 1m and L = 2m. What is the rate of change of the height of water in the tank? (in cm sec ) Answer iV. ANSWERS Answer i. Given:   y = x² − 5x ⇒   dy dx = 2x − 5 The x-coordinate increases at a constant rate of 0.2 units/s ⇒ dx dt = 0.2 Searched:     dy dt , for x = 4 Using the chain rule:   dy dx = dy dt · dt dx ⇒ 2x − 5 = dy dt · 1 0.2 therefore,      dy dt = (2x − 5)· 0.2 for x = 4,      dy dt = (2·4 − 5)· 0.2 = 3 · 0.2 = 0.6 The y-coordinate increases at a rate of 0.6 units per second. Answer ii. Given: y = 4 − 1 x   ⇒   dy dx = 1 x²   and   dy dt = 0.01 (y-coordinate rate of change) Searched:    dx dt ,  for x = 1 With the chain rule:     dy dx = dy dt · dt dx   ⇒   1 x² = 0.01 · dt dx therefore,   dt dx = 1 0.0x²   or   dx dt = 0.0x² 1 for x = 1,    dx dt = 0.01² 1 = 0.01 The x-coordinate increases at a rate of 0.01 units per second. Answer iii. Given: y = 1 2 − x   ⇒   dy dx = 1 (2 − x)² the y-coordinate is increasing at the constant rate of 0.5 units per second  ⇒  dy dt = 0.5 Searched:    dx dt ,  for x = 1 Assuming the chain rule:     dy dx = dy dt · dt dx   ⇒   1 (2 · x)² = 0.5 · dt dx therefore,   1 (2 − x)² · 0.5 = dt dx   or   dx dt = (2 − x)² · 0.5 1 for x = 1,    dx dt = 0.5 1 = 0.5 The x-coordinate increases at a rate of 0.5 units per second. Answer iV. Given: · The volume V of water in the tank by: V = w·L·H · The rate of change of the volume by: dV dt Searched: · The rate of change of the height H of water: dH dt V and H are functions of time. So let's differentiate: dV dt = W·L· dH dt , (with W and L as constants) therefore, dH dt = 1 W·L · dV dt Convert liters into cm3 and meters into cm as follows: 1 liter = 1 dcm3 = 1000 cm3 and 1 meter = 100 cm and, dH dt = 1 100cm·200cm · 20·1000cm3 = 1 cm sec The rate of change of the height H of water is 1 cm sec . TOP