In the previous section, we completed a discussion on the various applications of derivatives. In this section, we will see some miscellaneous examples.
Solved example 22.65
A car starts from a point P at time t = 0 seconds and stops at point Q. The distance x, in meters, covered by it, in t seconds is given by $\rm{x \,=\,t^2 \left(2 - \frac{t}{3} \right) }$.
Find the time taken by it to reach Q and also find the distance between P and Q.
Solution:
1. Consider the given equation: $\rm{x \,=\,t^2 \left(2 - \frac{t}{3} \right) \,=\,2t^2 - \frac{t^3}{3}}$
• This equation gives the total distance traveled when the reading in the stop-watch is 't'.
2. We know that for uniform motion, velocity = distance/time
• So for non-uniform motion, the instantaneous velocity at any instant 't' is given by dx/dt. Thus we get:
$\rm{v\,=\,\frac{dx}{dt}\,=\,4t - t^2}$
3. When the car reaches Q, it stops. So it's velocity at Q is zero.
• Equating the above equation to zero, we get:
4t - t2 = 0
⇒ t(4 - t) = 0
⇒ t = 0 and t = 4
• t = 0 is the instant at which the car is at P.
• t = 4 is the instant at which the car is at Q.
• So we can write:
The car takes 4 seconds to reach Q.
4. Using the given equation, we can write:
Distance traveled in the 4 seconds from P to Q =
$\rm{2(4)^2 - \frac{(4)^3}{3}\,=\,\frac{96 - 64}{3}\,=\,\frac{32}{3}~\text{m}}$
Solved example 22.66
A water tank has the shape of an inverted right circular cone with it's axis vertical and vertex lowermost. It's semi-vertical angle is tan−1 (0.5). Water is poured into it at a constant rate of 5 cubic meter per hour. Find the rate at which the level of water is rising at the instant when the depth of water in the tank is 4 m.
Solution:
1. In the fig.22.68 below,
• The water tank is drawn in red color. It is an inverted cone. The center of the circular base is O.
• The green dotted circle indicates the top surface of water.
♦ The center of this top surface of water, is C.
♦ Radius of this top surface of water, is r.
• The volume of water in the tank forms an inner inverted cone. Both cones have the same axis and the same vertex A.
 |
| Fig.22.68 |
• When the falling water forms the inner cone, three items will change continuously. They are:
♦ Volume of water (volume of inner cone) V
♦ Height of water (height of inner cone) h
♦ Radius of top surface of water (radius of the base of inner cone) r.
2. We have the formula for volume of a cone:
$\rm{V\,=\,\frac{1}{3} \pi r^2 h}$
• To eliminate r, we will use the semi-vertical angle.
Semi-vertical angle = ∠OAP = ∠CAB = tan−1 (0.5)
• So we can write:
$\rm{\tan (\angle CAB) \,=\,\frac{BC}{AC}\,=\,\frac{r}{h}}$
⇒ $\rm{\tan (\tan^{-1} (0.5)) \,=\,\frac{r}{h}}$
⇒ $\rm{0.5 \,=\,\frac{r}{h}}$
⇒ $\rm{r \,=\,\frac{h}{2}}$
• Now the expression for volume becomes:
$\rm{V\,=\,\frac{1}{3} \pi \left(\frac{h}{2} \right)^2 h}$
⇒ $\rm{V\,=\,\frac{1}{3} \pi \left(\frac{h^3}{4} \right)}$
(Note that, using the semi-vertical angle, we can eliminate either h or r. We must not eliminate h because, we want to find the rate of change of h)
3. Given that, water is being poured at a constant rate of 5 cubic meter per hour. So volume at any time 't' will be 5t, where 't' is in hours.
4. Equating the results in (2) and (3), we get:
$\rm{5t\,=\,\frac{1}{3} \pi \left(\frac{h^3}{4} \right)}$
⇒ $\rm{60t\,=\, \pi h^3}$
⇒ $\rm{h\,=\,\left(\frac{60t}{\pi}\right)^{1/3}}$
5. From this we get:
$\rm{\frac{dh}{dt}\,=\,\left(\frac{60}{\pi}\right)^{1/3} \left(\frac{1}{3} \right)t^{-2/3}}$
• This result will give the rate of change of h w.r.t time at any time 't'.
6. We want the rate at the time when the depth of water (height of inner cone) is 4 m. So we need to find that time.
• Using the result in (4), we can write:
$\rm{h\,=\,\left(\frac{60t}{\pi}\right)^{1/3}}$
⇒ $\rm{4\,=\,\left(\frac{60t}{\pi}\right)^{1/3}}$
⇒ $\rm{4^3 \,=\,\frac{60t}{\pi}}$
⇒ $\rm{64 \pi \,=\,60t}$
⇒ $\rm{t\,=\,\frac{64 \pi}{60}\,=\,\frac{16 \pi}{15}}$
7. Now we substitute this result in (5). We get:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\frac{dh}{dt}} & {~=~} &{\left(\frac{60}{\pi}\right)^{1/3} \left(\frac{1}{3} \right)t^{-2/3}} \\
{~\color{magenta} 2 } &{{\implies}} &{{\left.\frac{dh}{dt} \right|_{h=4}}} & {~=~} &{\left(\frac{60}{\pi}\right)^{1/3} \left(\frac{1}{3} \right)\left(\frac{16 \pi}{15} \right)^{-2/3}} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\left(\frac{(15)(4)}{\pi}\right)^{1/3} \left(\frac{1}{3} \right)\left(\frac{15}{16 \pi} \right)^{2/3}} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\left(\frac{(15)(4)}{\pi}\right)^{1/3} \left(\frac{1}{3} \right)\left(\frac{(15)(15)}{(16 \pi)(16 \pi)} \right)^{1/3}} \\
{~\color{magenta} 5 } &{{}} &{{}} & {~=~} &{\left(\frac{(15)(4)}{\pi}\right)^{1/3} \left(\frac{1}{3} \right)\left(\frac{(15)(15)}{4^3 (4) (\pi)(\pi)} \right)^{1/3}} \\
{~\color{magenta} 6 } &{{}} &{{}} & {~=~} &{\left(\frac{(15)}{\pi}\right)^{1/3} \left(\frac{1}{3} \right)\left(\frac{(15)(15)}{4^3 (\pi)(\pi)} \right)^{1/3}} \\
{~\color{magenta} 7 } &{{}} &{{}} & {~=~} &{\frac{15}{4} \left(\frac{1}{\pi}\right)^{1/3} \left(\frac{1}{3} \right)\left(\frac{1}{(\pi)(\pi)} \right)^{1/3}} \\
{~\color{magenta} 8 } &{{}} &{{}} & {~=~} &{\frac{5}{4} \left(\frac{1}{\pi}\right)^{1/3} \left(\frac{1}{(\pi)(\pi)} \right)^{1/3}} \\
{~\color{magenta} 9 } &{{}} &{{}} & {~=~} &{\frac{5}{4 \pi}~=~\frac{5}{4(22/7)}~=~\frac{35}{88}~\text{m/h}} \\
\end{array}$
Solved example 22.67
A man of height 2 meters walks at a uniform speed of 5 km/h away from a lamp post which is 6 meters high. Find the rate at which the length of his shadow increases.
Solution:
1. In the fig.22.69(a) below,
• AB is the lamp post. A is the position of the lamp. A ray of light from the lamp is shown as a yellow dotted line.
 |
| Fig.22.69 |
• CD is the initial position of the man just before he starts to walk. So BD is the initial distance of the man from the lamp post. It is denoted as d0.
• The ray of light passes just above C and creates the shadow of C at E. So DE is the initial length of the shadow. It is denoted as l0.
2. Let us assume that, in t hours, the man reaches the position C'D'. This is shown in fig.b.
• Then the distance DD' = 5t
• Also, length of shadow at that time will be D'E'. This length is denoted as l.
3. Consider the two triangles ABE' and C'D'E' in fig.b. These two triangles are similar. So we get:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\frac{AB}{BE'}} & {~=~} &{\frac{C'D'}{D'E'}} \\
{~\color{magenta} 2 } &{\implies} &{\frac{6}{BD + DD' + D'E'}} & {~=~} &{\frac{2}{l}} \\
{~\color{magenta} 3 } &{\implies} &{\frac{6}{d_0 + 5t + l}} & {~=~} &{\frac{2}{l}} \\
{~\color{magenta} 4 } &{\implies} &{\frac{d_0 + 5t + l}{6}} & {~=~} &{\frac{l}{2}} \\
{~\color{magenta} 5 } &{\implies} &{\frac{d_0 + 5t}{6}~+~\frac{l}{6}} & {~=~} &{\frac{l}{2}} \\
{~\color{magenta} 6 } &{\implies} &{\frac{d_0 + 5t}{6}} & {~=~} &{\frac{l}{3}} \\
{~\color{magenta} 7 } &{\implies} &{\frac{d_0 + 5t}{2}} & {~=~} &{l} \\
{~\color{magenta} 8 } &{\implies} &{l} & {~=~} &{\frac{d_0}{2}~+~\frac{5t}{2}} \\
{~\color{magenta} 9 } &{\implies} &{\frac{dl}{dt}} & {~=~} &{\frac{5}{2}~\text{km/h}} \\
\end{array}$
Solved example 22.68
The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base?
Solution:
1. In the fig.22.70 (a) below,
• ABC is an isosceles triangle. The base AB is fixed. Sides AC and BC have an initial length of l0 cm.
 |
| Fig.22.70 |
2. The sides are decreasing at the rate of 3 cm/s. So at the instant when the stop-watch reading is t, the length of both sides will be (l0 − 3t). This is shown in fig.b
3. In fig.b, the altitude (dashed yellow line) will have a length equal to $\rm{\sqrt{(l_0 - 3t)^2 ~-~(b/2)^2}}$
• This is the altitude at any instant t.
• So the area of the triangle at any instant t will be given by: $\rm{A = (1/2) (b) \sqrt{(l_0 - 3t)^2 ~-~(b/2)^2}}$
4. Now the rate of change of area can be obtained as:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{A} & {~=~} &{(1/2) (b) \sqrt{(l_0 - 3t)^2 ~-~(b/2)^2}} \\
{~\color{magenta} 2 } &{\implies} &{\frac{dA}{dt}} & {~=~} &{(1/2) (b) (1/2)((l_0 - 3t)^2 ~-~(b/2)^2)^{-1/2} 2 (l_0 – 3t)(-3)} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{(-3/2) (b)((l_0 - 3t)^2 ~-~(b/2)^2)^{-1/2} (l_0 – 3t)} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\frac{-3 b (l_0 – 3t)}{2 \sqrt{(l_0 - 3t)^2 ~-~(b/2)^2}}} \\
\end{array}$
• This result gives the rate of change at any instant t.
5. So we need the instant at which the sides become equal to base. For that, we use the result in (2):
Length of side at any instant = l0 − 3t
• So at the required instant, l0 − 3t = b
6. The above result can be used directly because (l0 − 3t) is available in the rate. We get:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\frac{dA}{dt}} & {~=~} &{\frac{-3 b (l_0 – 3t)}{2 \sqrt{(l_0 - 3t)^2 ~-~(b/2)^2}}} \\
{~\color{magenta} 2 } &{\implies} &{\left. \frac{dA}{dt} \right|_b} & {~=~} &{\frac{-3 b (b)}{2 \sqrt{(b)^2 ~-~(b/2)^2}}} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\frac{-3 b^2}{2 \sqrt{(3b^2)/4}}} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\frac{-3 b^2}{\sqrt{3b^2}}} \\
{~\color{magenta} 5 } &{{}} &{{}} & {~=~} &{-b \sqrt{3}~~\text{square cm per sec}} \\
\end{array}$
Solved example 22.69
A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic meter per hour. Then the depth of wheat is increasing at the rate of
(A) 1 m3/h (B) 0.1 m3/h (C) 1.1 m3/h (D) 0.5 m3/h
Solution:
1. Volume V of wheat in the cylinder, at t hours will be given by:
V = πr2h where r is the radius of the cylinder and h is the depth of wheat at t hours.
2. The above volume is equal to 314 t.
3. Equating the two results, we get: πr2h = 314 t
⇒ 3.14 r2h = 314 t
⇒ r2h = 100 t
⇒ 102h = 100 t
⇒ h = t
⇒ dh/dt = 1
• So the correct option is (A).
In the next section, we will see a few more miscellaneous examples.
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