4.6.1 Context
In this section we study basic dynamical systems that change with time. These systems can be modeled via the derivative. The key point is that the derivative measures the rate of change of a given quantity.
Observation It is important to note that a positive rate of change means that the quantity is increasing, while a negative rate of change indicates that the quantity is decreasing.
This follows from the Mean Value Theorem (MVT) (see Theorem 6 in 2.4.5).
Let \(f(t)\) be a differentiable quantity that changes with time \(t\), and assume that \(t_1 < t_2\), then (MVT) \[f(t_2)-f(t_1) = f'(t^*)(t_2-t_1), \;\;\; \text{ for some number } \;\; t^* \in (t_1,t_2).\] It follows that \vspace{.1in} if \(f'(t) > 0 \) for all \(t\), then \(f(t_2)-f(t_1) > 0\), that is, \(f(t_2) > f(t_1)\) and \(f\) is increasing;
if \(f'(t) < 0 \) for all \(t\), then \(f(t_2)-f(t_1) < 0\), that is, \(f(t_2) < f(t_1)\) and \(f\) is decreasing.
Example 1
The edges of an ice-cube decrease at a rate of 2 mm/min, at what rate is the volume of the cube changing when the edges measure 12 mm?
Solution Let \(x = x(t)=\) length of each of the edges of the ice-cube. Then, \(x'(t) = -2\) mm/min.
Since \(V= x^3\), to compute the rate of change of \(V\) with respect to time, we can apply the chain rule as follows:
\[V= x^3 = (x(t))^3 \;\; \Longrightarrow\;\;\frac{d}{dt}V= \frac{d}{dt}(x(t))^3 = 3(x(t))^2\cdot x'(t).\]
We can now replace the data provided in the statement of the problem: \(x'(t) = -2\) mm/min, and the wanted rate is for the time \(t\) when the edges measure \(x(t)=12\) mm, thus
\[V’ = 3(12)^2(-2) \text{mm}^3/\text{min},\; \text{ that is, }\; V’=-864 \text{mm}^3/\text{min.}\] Negative, because the volume is decreasing as the ice-cube melts.
Example 2
A snowball melts at a rate of 75 mm\(^3\)/min, at what rate is its radius changing when the volume is \(1125\pi\) mm\(^3\)?}
\[V= \frac{4}{3}\pi r^3 \;\; \underset{?}{\Longrightarrow} \;\; V’ = 4\pi r^2r’.\]
When \( V= 1125\pi \) mm\(^3\), then the magnitude of the radius is obtained as follows:
\[1125\pi = \frac{4}{3}\pi r^3 \;\; \Longrightarrow \;\; r=\frac{15}{\sqrt[3]{4}}, \]
It now follows that
\[V’ = 4\pi r^2r’ \;\; \Longrightarrow\;\; -75 = 4\pi \left (\frac{15}{\sqrt[3]{4}} \right )^2r’.\]
Hence
\[r’=-\frac{ \sqrt[3]{2}}{6\pi \,}\;\frac{\text{mm}}{\text{min}},\]
when the volume of the snowball is \(1125\pi\) mm\(^3\).
4.6.2 Exercises
- The edges of a square increase at a rate of \(2/3\) cm/min. At what rate is the area of the square increasing when the edges measure \(8\) cm?
- The radius of a circle increases at a rate of \(1/2\) cm/min. At what rate is the area of the circle increasing when the radius is \(4\) cm?
- The radius of a spherical balloon increases at a rate of \(2\) cm/min. At what rate is the volume of the balloon increasing when the radius is \(10\) cm?
- The radius of a cylinder changes at a rate of \(2/3\) cm/min and its height at a rate of \(-3/4\) cm/min. At what rate is the volume of the cylinder changing when the radius is \(6\) cm and the height is 9 cm? Is the volume increasing or decreasing?
- An ice-cube melts at a rate of 3 cm\(^3\) per minute. At what rate are its edges changing when its volume is 27 cm\(^3\)?
- A particle follows a trajectory given by the equation. The rate of change, at a certain position, is given for one coordinate. Find the rate of change of the other coordinate,
- \(\displaystyle x^2+ \frac{y^2}{4} = 2\), \(x’=2\), at \(( 1,2)\).
- \(\displaystyle x^2 -4 x y + y^2 = 1\), \(y’=-1\), at \(( 1,2)\).
- \(\sin (\pi xy^2) = x^2-y^2\), \(x’=0\), at \(( 1,-1)\).
- \(\displaystyle \tan \frac{\pi\sqrt{xy}}{4} =xy^3\), \(y’ = 1\), at \(( -1 , -1)\).
Outline of Steps to Solve Related Rates Problems
When solving related rates problems, it is convenient to follow a systematic procedure, which we outline here:
Step 1. Begin by identifying the quantities whose rates are known and unknown, and the known rates. Introduce notation that is closely related to these quantities. If possible, draw a picture labeling all relevant information.
Step 2. Find a relation between the various quantities.
Step 3. Differentiate with respect to \(\pmb{t}\) to relate the rates of the quantities that change. Here you will need the Chain Rule.
Step 4. Substitute the given particular values for the quantities that change into the equation found in Step 4.
4.6.3 Additional Examples
Example 1 Point \(A\) is located at \((0,5)\), while point \(B\) moves upward along the \(y\)-axis at a rate of \(2\) units/m.
(a) At what rate is the distance between points \(A\) and \(B\) increasing at the time when point \(B\) is located at \((0,12)\)?
(b) At what rate is the angle \(\angle OAB\) increasing, at the same time as in (a)?
Solution (a)
Step 1. The quantity whose rate is unknown is the distance between \(A\) and \(B\). Let \(s =s(t)\) be this distance at time \(t\). Let \( y = y(t)\) be the \(y\)-coordinate of \(B\) at time \(t\) (see Figure 4.6)
Figure 4.6
Step 2. Relations among the different quantities:
\[ s^2 = y^2+5^2\]
Step 3. Differentiating with respect to time and using the Chain Rule:
\[2ss’ = 2yy’\]
Step 4. Substitute the given data into the equations relating the rates.
Since \(y=12\) and \(y’ =2\), then
\[s= \sqrt{12^2+5^2}=\sqrt{169} = 13 \;\; \Longrightarrow \;\; s’= \frac{y}{s}y’ = \frac{12}{13}\cdot 2=\frac{24}{13} \; \left (\frac{\text{units}}{\text{min}}\right ).\]
Solution (b)
Step 1. Let \(\theta =\theta(t) = \angle OAB\). This is the quantity whose rate is unknown (see Figure 4.6).
Step 2. Relation among the changing quantities.
\[\tan \theta = \frac{y}{5}\]
Step 3. Differentiating with respect to \(t\).
\[\sec^2 \theta \cdot \theta’= \frac{1}{5}y’ \;\; \Longrightarrow \;\; \theta’ = \frac{1}{5}\cdot\frac{1}{\sec^2\theta} \cdot y’.\]
Step 4. Substitute the given data into the equations relating the rates.
Since \(y=12\) and \(y’=2\), then
\[\tan \theta = \frac{12}{5}\;\; \Longrightarrow\;\; \sec^2\theta = \tan^2\theta + 1 = \frac{169}{25}\;\;\Longrightarrow \;\;\theta’ = \frac{1}{5} \cdot \frac{25}{169} \cdot 2= \frac{10}{169}\; \left (\frac{\text{rad}}{\text{min}}\right ).\]
Example 2
A reservoir with the shape of a right circular-cone has a base of radius 2 m and an altitude of 4 m. Water is being pumped into the tank a rate of \(\displaystyle \frac{1}{2}\) m\(^3\) per minute. At what rate is the level of the water rising when the depth of the water is 3 m, if (a) the vertex of the tank is at the top; (b) the cone is inverted, that is, the vertex of the tank is at the bottom.
Figure 4.7 (a)
Figure 4.7 (b)
Solution (a)
Step 1. Quantities that change as time \(t\) changes and their rates, and a diagram (see Figure 4.7(a)).
Let \(h = h(t) = \) depth of the water, \(r = r(t) = \) radius of the surface of the water, and \(V=V(t) =\) volume of water in the tank at time \(t\). Here, \(\displaystyle V’= \frac{1}{2}\) m\(^3\)/min.
Step 2. Relation between the quantities that change.
The volume of a conical frustum with radii \(r_1\) and \(r_2\), and height \(h\) is
\(\displaystyle V=\frac{1}{3}h\left (r_1^2+ r_1 r_2 + r_2^2\right )\). Thus,
the volume of water in the tank is given by
\[V= \frac{1}{3}h\left (2 ^2+ 2 r+ r^2\right ).\]
Using the similarity of the two triangles in Figure 4.7(a), we have
\[\frac{4}{2} = \frac{4-h}{r} \;\; \Longleftrightarrow \;\; r = \frac{1}{2}(4-h).\]
\[V= \frac{1}{3}h\left (4+ (4-h)+ \left (\frac{1}{2}(4-h)\right )^2\right ) =4 h – h^2 +\frac{h^3}{12} .\]
Step 3. Related rates:
\[V’ = 4h’-2hh’+\frac{1}{4}h^2h’.\]
Step 4. Substitute the given data.
\[\frac{1}{2}= 4h’-2\cdot 3 h’+\frac{1}{4}3^2h’ \Longleftrightarrow \frac{1}{2} = \frac{1}{4}h’\Longleftrightarrow h’=2.\]
Thus, the rate at which the water is rising when \(h=3\) m is 2 m/min.
Solution (b) Here is an outline of the solution. Justify each step providing all necessary details.
Step 1. Same as in (a) (see Figure 4.7(b)).
Step 2. In this case,
\[\frac{r}{h}=\frac{2}{4}\;\; \underset{?}{\Longrightarrow} \;\; V=\frac{1}{6}\pi h^3.\]
Step 3. Applying the Chain Rule: \(\displaystyle V’ = \frac{1}{2}\pi h^2h’\).
Step 4. When \(h =3\) m, it follows that \(\displaystyle h’\underset{?}{=}\frac{1}{9\pi}\) m/min.
Example 3
Point \(A\) revolves counterclockwise around the circle of radius 2 m with center at the origin at a rate of 3 rev/min, while point \(B\) located at \((5,0)\) remains stationary. At what rate is the distance between \(A\) and \(B\) changing when \(A\) is located at \((\sqrt{3},1)\)?
Solution.
When solving this example, there are two points to keep in mind:
[-] The statement that \(A\) revolves counterclockwise around the origin at a rate of 2 revolutions per minute translates as
\[ 2\, \frac{\text{rev}}{\text{min}} = 2 \cdot 2\pi\, \frac{\text{rad}}{\text{min}}.\]
[-] The Law of Cosines. Consider a triangle \(ABC\), with \(\angle A = \theta \), \(|BC| = a\), \(|AC| = b\), and \(|AB| = c\) (see Figure 4.8(a)), then
\[c^2 = a^2 + b^2 – 2ab\cos \theta .\]
Figure 4.8
Figure 4.8
Step 1. Quantities that change as time \(t\) changes and their rates; a diagram (see Figure 4.8(b)).
Let \(\theta =\theta(t)\) angle of revolution and \(s=s(t) =\) distance between \(A\) and \(B\) at time \(t\). We have \(\theta’=\) 3 rev/min \(= 6\pi\) rad/min
Step 2. Relation between the quantities that change:
\[s^2= 2^2+5^2-2(2)(5)\cos \theta\;\;\;\text{(Law of Cosines).}\]
Step 3. Related rates:
\[2s\, s’ = 20\sin\theta \,\theta’\,.\]
Step 4. Substitute the given data.
When \(A\) is at \((\sqrt{3},1)\), \(\displaystyle \theta = \frac{\pi}{6}\). Then
\[ s^2= 2^2+5^2-2(2)(5)\cos \frac{\pi}{6} =29-20\left(\frac{\sqrt{3}}{2}\right)=29-10\sqrt{3}\Longrightarrow s=\sqrt{29-10\sqrt{3}}\Longrightarrow\]
\[\Longrightarrow s’ = \frac{\displaystyle 20\left(\sin\frac{\pi}{6}\right)(6\pi)}{2s}
=\frac{\displaystyle 20\left(\frac{1}{2}\right)(6\pi)}{2\sqrt{29-10\sqrt{3}}}
= \frac{30\pi}{\sqrt{29-10\sqrt{3}}} \; \left (\frac{\text{m}}{\text{min}}\right ).\]
Example 4
A 25-ft ladder leans against a wall. The top of the ladder slides down the wall at a rate of 2 ft/sec, (a) how fast is the bottom moving along the ground when the bottom of the ladder is 7 ft from the wall? (b) at what rate is the acute angle formed by the ground and the ladder changing at the same time as in (a)?
Solution (a)
Step 1. Quantities that change as time \(t\) changes, their rates, and diagram (see Figure 4.7(a)).
Let \(x= x(t) =\) distance from the base of the ladder to the wall, \(y= y(t) =\) distance from the top of the ladder to the ground at time \(t\). Here, \(y’=-2\) ft/sec (decreasing).
Figure 4.9
Step 2. Relations among the different quantities:
\[25^2 = x^2+y^2.\]
Step 3. Differentiating with respect to time and using the Chain Rule:
\[2xx’ + 2yy’ = 0.\]
Step 4. Substitute the given data into the equations relating the rates.
Since \(x=7\), then
\[y= \sqrt{25^2-7^2}=\sqrt{576} = 24 \;\; \Longrightarrow \;\; x’= -\frac{y}{x}y’ = \frac{24}{7}\cdot(- 2)=-\frac{48}{7} \; \left (,\frac{\text{ft}}{\text{sec}}\right ).\]
Solution (b)
Step 1. Let \(\theta =\theta(t) =\) acute angle that the ladder makes with the ground at time \(t\).
Step 2. Relations among the different quantities:
\[\sin \theta = \frac{y}{25}.\]
Step 3. Differentiating with respect to time and using the Chain Rule:
\[\cos \theta \cdot \theta’= \frac{1}{25}y’.\]
Step 4. Substitute the given data into the equations relating the rates.
Since \(x=7\) and \(y’=-2\), then
\[\cos \theta = \frac{7}{25}\;\; \Longrightarrow \;\;\theta’ = \frac{1}{25}\cdot (-2) \cdot \frac{25}{7} =- \frac{2}{7}\; \left (\frac{\text{rad}}{\text{sec}}\right ).\]
4.6.4 Exercises
- A 25-ft ladder is leaning against a wall. If we push the ladder toward the wall at a rate of 1 ft/sec, and the bottom of the ladder is initially 20 ft away from the wall, (a) how fast does the ladder move up the wall 5 sec after we start pushing? (b) How fast is the acute angle made by the ground and the ladder increasing 5 sec after we start pushing?
- Two airplanes are flying in the air at the same height: airplane \(A\) is flying east at 250 mi/h and airplane \(B\) is flying north at 300 mi/h. If they are both heading to the same airport \(O\),
(a) at what rate is the distance between the airplanes changing when \(O\) is located 30 miles east of airplane \(A\) and 40 miles north of airplane \(B\) ?
(b) At what rate is the angle \(\angle OAB\) changing at the same instant as in (a)?
- A 6-ft-tall person walks away from a 10-ft lamppost at a constant rate of 3 ft/sec. (a) What is the rate that the tip of the shadow moves away from the pole when the person is 10 ft away from the pole? (b) What is the rate that the angle formed by the base of the lamppost the tip of the shadow and the top of the lamppost changes when the person is 10 ft away from the pole?
- A helicopter starting on the ground is rising directly into the air at a rate of 25 ft/sec. You are running on the ground starting directly under the helicopter at a rate of 10 ft/sec. (a) Find the rate of change of the distance between the helicopter and yourself after 5 sec. (b) What is the rate at which the distance between you and the helicopter is changing when the helicopter has risen to a height of 60 ft in the air, assuming that, initially, it was 30 ft above you?
- The radius of a sphere is increasing at a rate of 9 cm/sec. Find the radius of the sphere when the volume and the radius of the sphere are increasing at the same numerical rate.
- The base of a triangle is shrinking at a rate of 1 cm/min and the height of the triangle is increasing at a rate of 5 cm/min. Find the rate at which the area of the triangle changes when the height is 22 cm and the base is 10 cm.
- A triangle has two constant sides of length 3 ft and 5 ft. The angle between these two sides is increasing at a rate of 0.1 rad/sec. Find the rate at which the area of the triangle is changing when the angle between the two sides is \(\displaystyle \frac{\pi}{6}\).
- Consider a right circular cone that is leaking water. The dimensions of the conical tank are a height of 16 ft and a radius of 5 ft.
- How fast does the depth of the water change when the water is 10 ft high if the cone leaks water at a rate of 10 ft\(^3\)/min?
- Find the rate at which the surface area of the water changes when the water is 10 ft high if the cone leaks water at a rate of 10 ft\(^3\)/min.
- If the water level is decreasing at a rate of 3 in/min when the depth of the water is 8 ft, determine the rate at which water is leaking out of the cone.
- A trough has ends shaped like isosceles triangles, with width 3 m and height 4 m, and the trough is 10 m long. Water is being pumped into the trough at a rate of 5 m\(^3\) /min. At what rate does the height of the water change when the water is 1 m deep?
- A tank is shaped like an upside-down square pyramid, with base of 4 m by 4 m and a height of 12 m. How fast does the height increase when the water is 2 m deep if water is being pumped in at a rate of \(\displaystyle \frac{2}{3}\) m/sec?
- Gravel is being unloaded from a truck and falls into a pile shaped like a cone. The radius of the cone base is three times the height of the cone.
- If the gravel is unloaded at a rate of 10 ft\(^3\)/min, find the rate at which the height of the gravel changes when the pile has a height of 5 ft.
- If the pile is 5 ft high and the height is increasing at a rate of 4 in/min, find the rate at which the gravel is being unloaded.
- You are stationary on the ground and are watching a bird fly horizontally at a rate of 10 m/sec. The bird is located 40 m above your head. How fast does the angle of elevation change when the horizontal distance between you and the bird is 9 m?
- You stand 40 ft from a bottle rocket on the ground and watch as it takes off vertically into the air at a rate of 20 ft/sec. Find the rate at which the angle of elevation changes when the rocket is 30 ft in the air.
- A lighthouse, L, is on an island 4 mi away from the closest point, P, on the beach. If the lighthouse light rotates clockwise at a constant rate of 10 revolutions/min, how fast does the beam of light move across the beach 2 mi away from the closest point on the beach?
- You are walking to a bus stop at a right-angle corner. You move north at a rate of 2 m/sec and are 20 m south of the intersection. The bus travels west at a rate of 10 m/sec away from the intersection – you have missed the bus! What is the rate at which the angle between you and the bus is changing when you are 20 m south of the intersection and the bus is 10 m west of the intersection?
- Point \(A\) moves upward, starting at the origin, along the line \(y=x\) (in the first quadrant) at a rate of 3 cm/sec, while point \(B\) remains stationary at \((5,0)\). At what rate is the distance between \(A\) and \(B\) changing when \(A\) is located at \((2,2)\)?
- Point \(A\) moves upward, starting at the origin, along the line \(y=-\sqrt{3} x\) (in the second quadrant) at a rate of 2 cm/sec, while point \(B\) starts moving, starting from the origin, 10 seconds later along the positive \(x\)-axis at a rate pf 3 cm/sec. At what rate is the distance between \(A\) and \(B\) changing when \(B\) has traveled for 5 seconds?
- An isosceles trapezoid is inscribed in a semicircle so that one base is the diameter, and the second base is half the diameter. If the radius of the semicircle increases at a rate of 2 cm/min, find the rate at which the area of the trapezoid changes when the radius is 10 cm.