KindiakMath

Rates of Change

Definition 1. Define the rate of change of a quantity x with respect to time t by \displaystyle \frac{\mathrm dx}{\mathrm dt}.

Problem 1. A latter of length 10 m rests with its foot on horizontal ground and its top against a smooth vertical wall. The foot of the ladder slides away from the wall at a constant rate of 0.2 m/s.

Let x m denote the distance of the foot from the wall and y m denote the height of the ladder above the ground.

Show that

\displaystyle \frac{\mathrm dy}{\mathrm dt} = -\frac xy \cdot \frac{\mathrm dx}{\mathrm dt}.

Deduce the rate at which the ladder is falling down along the wall at x = 6.

(Click for Solution)

Solution. By Pythagoras’ theorem,

x^2 + y^2 = 10^2.

Differentiating all sides with respect to t,

\displaystyle \frac{\mathrm d}{\mathrm dt}(x^2 + y^2) = \frac{\mathrm d}{\mathrm dt}(10^2).

On the left-hand side, the chain rule yields

\begin{aligned} \frac{\mathrm d}{\mathrm dt}(x^2 + y^2) &= \frac{\mathrm d}{\mathrm dt}(x^2) + \frac{\mathrm d}{\mathrm dt}(y^2) \\ &= 2x \cdot \frac{\mathrm dx}{\mathrm dt} + 2y \cdot \frac{\mathrm dy}{\mathrm dt} \\ &= 2 \left( x \cdot \frac{\mathrm dx}{\mathrm dt} + y \cdot \frac{\mathrm dy}{\mathrm dt} \right). \end{aligned}

On the right-hand side, \displaystyle \frac{\mathrm d}{\mathrm dt}(10^2) = 0. Substituting all displays,

\begin{aligned} 2 \left( x \cdot \frac{\mathrm dx}{\mathrm dt} + y \cdot \frac{\mathrm dy}{\mathrm dt} \right) &= 0 \\ x \cdot \frac{\mathrm dx}{\mathrm dt} + y \cdot \frac{\mathrm dy}{\mathrm dt} &= 0 \\ y \cdot \frac{\mathrm dy}{\mathrm dt} &= -x \cdot \frac{\mathrm dx}{\mathrm dt} \\ \frac{\mathrm dy}{\mathrm dt} &= -\frac xy \cdot \frac{\mathrm dx}{\mathrm dt}. \end{aligned}

At x = 6,

6^2 + y^2 = 10^2 \quad \Rightarrow \quad y = 8.

Since \displaystyle \frac{\mathrm dx}{\mathrm dt} = 0.4,

\displaystyle \frac{\mathrm dy}{\mathrm dt} = -\frac xy \cdot \frac{\mathrm dx}{\mathrm dt} = -\frac 68 \cdot 0.4 = -0.3\, \text{m/s}.

Therefore, the ladder is falling down along the wall at a rate of 0.3 m/s.

Problem 2. Car A travels south toward a junction, and Car B travels east away from the junction. Let s km denote the distance between Car A and Car B.

Given that Car A travels at a speed of 60 km/h and Car B travels at a speed of 80 km/h, determine the rate of change of s at the point in time where Car A is located 3 km north of the junction, while Car B is located 4 km east of the junction.

(Click for Solution)

Solution. By Pythagoras’ theorem,

x^2 + y^2 = s^2.

Differentiating on both sides similar to Problem 1,

\displaystyle 2x \cdot \frac{\mathrm dx}{\mathrm dt} + 2y \cdot \frac{\mathrm dy}{\mathrm dt} = 2s \cdot \frac{\mathrm ds}{\mathrm dt}.

At x = 4, y = 3, s = 5. Since \displaystyle \frac{\mathrm dx}{\mathrm dt} = 80 and \frac{\mathrm dy}{\mathrm dt} = -60, substituting the values yields

\displaystyle 2 \cdot 4 \cdot 80 + 2 \cdot 3 \cdot (-60) = 2 \cdot 5 \cdot \frac{\mathrm ds}{\mathrm dt}.

Solving the equation yields \displaystyle \frac{ \mathrm ds }{ \mathrm dt } = 28\, \text{km}/\text{h}.

Problem 3. A cup of coffee is placed in a room where the ambient temperature is 20°C. According to Newton’s law of cooling, the temperature T°C of the coffee at time t minutes after it is placed in the room satisfies the equation

T = 20 + 80e –0.05t .

Show that the rate of change of T is directly proportional to (T – 20).

(Click for Solution)

Solution. Differentiating with respect to time,

\displaystyle \frac{ \mathrm dT }{ \mathrm dt } = 80 \cdot (-0.05) e^{-0.05t}.

On the other hand,

\displaystyle e^{-0.05t} = \frac{T - 20}{80}.

Substituting the display,

\displaystyle \frac{ \mathrm dT }{ \mathrm dt } = 80 \cdot (-0.05) \cdot \frac{T - 20}{80} = -0.05 \cdot (T-20).

Therefore, \displaystyle \frac{\mathrm dT}{\mathrm dt} \propto (T-20).

Problem 4. The displacement x cm from a fixed point O at time t seconds of an object moving in a straight line is given by

x = 3 sin(2t).

Define

  • the velocity, v cm/s, of the object as the rate of change of the displacement of the object,
  • the acceleration, a cm/s2, of the object as the rate of change of the velocity of the object.

Show that a is directly proportional to x.

(Click for Solution)

Solution. Differentiating twice with respect to time,

\begin{aligned} v &= \frac{\mathrm ds}{\mathrm dt} = 3 \cdot \cos(2t) \cdot 2 = 6 \cos(2t), \\ a &= \frac{\mathrm dv}{\mathrm dt} = 6 \cdot (-{\sin(2t)} \cdot 2) = -12 \sin(2t). \end{aligned}

Since \sin(2t) = x/3, substituting into the display yields a = -4x. Therefore, a \propto x.

—Joel Kindiak, 7 Apr 26, 2159H

,

Published by

joelkindiak

Leave a comment