In the previous section, we completed a discussion on continuity and differentiability. We saw how to find the derivatives of various functions.
Derivatives have many applications in science, engineering, economics etc., In this chapter, we will see some of those applications.
First we will see how derivatives help us to find the rate of change of quantities. Let us see an example. It can be written in 6 steps:
1. An object is moving in a straight line.
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It started with an initial velocity zero. That is, it started from rest.
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But it has a constant acceleration 'a'.
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A stop-watch is turned on at the instant when the object starts it's motion.
2. In such a situation, the distance traveled by the object when the stop-watch reading is 't', can be obtained using the formula: $\rm{s \,=\, \frac{1}{2} a t^2 }$
3. Using this formula, we can plot a graph with t along the x-axis and s along the y-axis. If the acceleration 'a' is $\rm{3 ~ m s^{-2}}$, the graph will be as shown in fig.22.1 below:
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| Fig.22.1 |
4. Since the object is moving with a constant acceleration 'a', the velocity will be changing continuously. That is, velocity is not uniform.
5. We can find the velocity at any instant by using the derivative $\rm{\frac{ds}{dt}}$ (recall that velocity is the rate of change of distance).
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In our present case, $\rm{s \,=\, f(t)\,=\,\frac{1}{2}. 3. t^2 \,=\, 1.5 t^2}$
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So $\rm{f'(t) \,=\, \frac{ds}{dt} \,=\, \frac{d}{dt} \left(1.5 t^2 \right) \,=\, 1.5 \times 2t = 3t}$
6. Let us see an example:
The velocity when the stop-watch reading is 0.3 s, is (3 × 0.3) = 0.9 $\rm{m s^{-1}}$
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From the graph, we see that:
Rate of change (velocity in this case) is same as slope of the tangent.
Let us see some solved examples:
Solved example 22.1
If the radius of a circle is changing, what is the rate of change of the area of that circle, w.r.t radius?
Solution:
1. The area of circle is given by the equation: $\rm{A~=~\pi r^2}$
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Where 'A' is the area and 'r' is the radius.
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So it is clear that, if the radius changes, the area will also change.
2. Using the above equation, we can draw a graph which shows the variation of area. It is given in fig.22.2 below:
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| Fig.22.2 |
3. We see that, the graph is a curve. That means, the rate of change of area is not uniform. If the rate of change was uniform, we would have obtained a straight line graph.
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Due to the non-uniform nature, we can find only the instantaneous rate of change $\rm{\frac{dA}{dr}}$
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So $\rm{f'(r) \,=\, \frac{dA}{dr} \,=\, \frac{d}{dr} \left(\pi r^2 \right) \,=\, \pi \times 2r = 2 \pi r}$
4. Let us see an example:
The rate of change of area when the radius is 0.2 m is:
(2 × π × 0.2) = 1.26 $\rm{m^2 m^{-1}}$
• That means, for every one meter change of radius, the area changes by 1.26 m2.
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From the graph, we see that:
Rate of change of area is same as slope of the tangent at (r=0.2).
• The instantaneous rate of change of y w.r.t x, at a particular point x0, is denoted as: $\rm{\left. \frac{dy}{dx} \right|_{x =x_0}}$
• It can be also denoted as $\rm{f'(x_0)}$
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So in the above example, we can write:
$\rm{\left. \frac{dA}{dr} \right|_{r = 0.2}~=~1.26 ~ m^2 m^{-1}}$
Solved example 22.2
The radius of a circle is increasing at the rate of 4 cm s−1. what is the rate of change of the area of that circle, w.r.t time, at the instant when the radius is 10 cm?
Solution:
1. The area of circle is given by the equation: $\rm{A~=~\pi r^2}$
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Where 'A' is the area and 'r' is the radius.
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So it is clear that, if the radius changes, the area will also change.
2. The radius is increasing continuously. So the area will also increase continuously.
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In the previous example, we saw that, the instantaneous rate of change of area w.r.t radius is $\rm{\frac{dA}{dr}}$.
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In this example, we are asked to find the instantaneous rate of change of area w.r.t time, which is $\rm{\frac{dA}{dt}}$.
3. So we need to bring 't' also into the calculations.
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The rate of increase of radius is given as 4 cm s−1. We can write:
♦ When t = 0, r = 0
♦ When t = 1, r = 4 cm
♦ When t = 2, r = 8 cm
♦ When t = 3, r = 12 cm
so on . . .
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That means, at any time t, the radius will be 4t.
4. Now we can modify the equation for area:
$\rm{A~=~\pi r^2 ~=~ \pi (4t)^2}$
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This gives: $\rm{\frac{dA}{dt} ~=~16 \pi \times 2t ~=~32 \pi t}$
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Using this result, we can calculate the rate of increase of area w.r.t time, at any instant.
5. We are asked to calculate the rate of increase of area w.r.t time, at the instant when radius is 10 cm.
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So we first need to find that instant.
♦ We have: r = 4t = 10 cm.
♦ So t = 10/4 = 2.5 s
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That means, the radius will be 10 cm when t = 2.5 s.
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Substituting this t in (4), we get:
$\rm{\frac{dA}{dt} ~=~32 \pi t = 32 \times \pi \times 2.5 ~=~80 \pi ~cm^2 s^{-1}}$
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That is., $\rm{\left. \frac{dA}{dt} \right|_{r = 10}~=~80 \pi ~ cm^2 s^{-1}}$
• We can say that:
At the instant when radius is 10 cm, the area is increasing at the rate of $\rm{80 \pi ~ cm^2}$ per second.
At any other instant, the rate will be different.
Solved example 22.3
The volume of a cube is increasing at the rate of 9 cm3 s−1. what is the rate of change of the surface area of that cube, w.r.t time, at the instant when the length of an edge is 10 cm?
Solution:
1. Let V be the volume, S the surface area and l the length of edge.
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We can easily calculate $\rm{\frac{dS}{dl}}$ because, S = 6l2.
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But we are asked to find $\rm{\frac{dS}{dt}}$
2. So we need to bring 't' also into the calculations.
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The rate of increase of volume is given as 9 cm3 s−1. We can write:
♦ When t = 0, V = 0 cm3.
♦ When t = 1, V = 9 cm3.
♦ When t = 2, V = 18 cm3.
♦ When t = 3, V = 27 cm3.
so on . . .
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That means, at any time t, the volume will be 9t.
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That means, at any time t, the length of an edge l can be obtained using the equation: $\rm{l~=~(V)^{\frac{1}{3}}~=~(9t)^{\frac{1}{3}}}$.
3. So at any time t, the surface area S can be obtained using the equation:
$\rm{S ~=~6 \times l^2~=~6 \left[(9t)^{\frac{1}{3}} \right]^2~=~6 \left[9^{\frac{2}{3}} t^{\frac{2}{3}} \right] }$
4. Then the rate of change of S w.r.t to time can be obtained using the equation:
$\rm{\frac{dS}{dt}\,=\,6 \times 9^{\frac{2}{3}} \times \frac{2}{3} \times t^{\frac{-1}{3}}}$
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Using this result, we can calculate the rate of increase of surface area w.r.t time, at any instant.
5. We are asked to calculate the rate of increase of surface area w.r.t time, at the instant when length of edge is 10 cm.
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So we first need to find that instant.
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{l} & {~=~} &{(9t)^{\frac{1}{3}}} \\
{~\color{magenta} 2 } &{\implies} &{10} & {~=~} &{(9t)^{\frac{1}{3}}} \\
{~\color{magenta} 3 } &{\implies} &{10^3} & {~=~} &{9t} \\
{~\color{magenta} 4 } &{\implies} &{t} & {~=~} &{\frac{1000}{9}} \\
\end{array}$
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That means, the length of edge will be 10 cm when t = 1000/9 s.
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Substituting this t in (4), we get:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\frac{dS}{dt}} & {~=~} &{6 \times 9^{\frac{2}{3}} \times \frac{2}{3} \times t^{\frac{-1}{3}}} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{6 \times 9^{\frac{2}{3}} \times \frac{2}{3} \times \left({\frac{1000}{9}} \right)^{\frac{-1}{3}}} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{2 \times 9^{\frac{2}{3}} \times 2 \times \left({\frac{9}{1000}} \right)^{\frac{1}{3}}} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{2 \times 9 \times 2 \times \left({\frac{1}{10}} \right)} \\
{~\color{magenta} 5 } &{{}} &{{}} & {~=~} &{\frac{36}{10}} \\
{~\color{magenta} 6 } &{{}} &{{}} & {~=~} &{3.6~ \rm{cm^2 \,s^{-1}}} \\
\end{array}$
• That is., $\rm{\left. \frac{dS}{dt} \right|_{l = 10}~=~3.6 ~ cm^2 s^{-1}}$
• We can say that:
At the instant when length of edge is 10 cm, the surface area is increasing at the rate of $\rm{3.6 ~ cm^2}$ per second.
At any other instant, the rate will be different.
Solved example 22.4
The length x of a rectangle is decreasing at the rate of 3 cm / minute and width y is increasing at the rate of 2 cm / minute. When x = 10 cm and y = 6 cm, find the rates of changes of (a) the perimeter and (b) area of the rectangle.
Solution:
Part (a):
1. Suppose that, the initial length is x0.
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Given that, x is decreasing at the rate of 3 cm / minute. So we can write:
♦ When t = 1, x = x0 − (3 × 1)
♦ When t = 2, x = x0 − (3 × 2)
♦ When t = 3, x = x0 − (3 × 3)
so on . . .
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Thus, the length at any time 't' is given by:
x = x0 − (3 × t)
2. Suppose that, the initial width is y0.
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Given that, y is increasing at the rate of 2 cm / minute. So we can write:
♦ When t = 1, y = y0 + (2 × 1)
♦ When t = 2, y = y0 + (2 × 2)
♦ When t = 3, y = y0 + (2 × 3)
so on . . .
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Thus, the width at any time 't' is given by:
y = y0 + (2 × t)
3. Perimeter at any time 't' will be given by:
P= 2[x+y] = 2[(x0 − 3t) + (y0 + 2t)] = 2[x0 + y0 − t] = 2(x0 + y0)− 2t
4. Rate of change of perimeter w.r.t time, at any instant, will be given by:
$\rm{\frac{dP}{dt}\,=\, \frac{d}{dt} \left[2(x_0 + y_0) - 2t \right]\,=\,-2}$
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We see that, the rate is a constant. That means, the rate is independent of x, y or t.
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We are asked to find the rate when x = 10 cm and y = 6 cm. At this length and breadth also, the rate will be −2 cm/minute.
• We can say that:
At the instant when x = 10 cm and y = 6 cm, the perimeter is decreasing at the rate of $\rm{2 ~ cm}$ per minute.
At any other instant, the rate will be same.
Part (b):
1. Area at any time 't' will be given by:
$\rm{A\,=\, xy \,=\,(x_0 - 3t)(y_0 + 2t)\,=\,(x_0 y_0 + 2 x_0 t - 3 y_0 t - 6 t^2)}$
⇒ $\rm{A\,=\,x_0 y_0 + (2 x_0 - 3 y_0) t - 6 t^2}$
2. Rate of change of area w.r.t time, at any instant, will be given by:
$\rm{\frac{dA}{dt}\,=\, \frac{d}{dt} \left[x_0 y_0 + (2 x_0 - 3 y_0) t - 6 t^2 \right]\,=\,2 x_0 - 3 y_0 - 12 t}$
3. We are asked to find the dA/dt at the instant when x = 10 cm and y = 6 cm.
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So we first need to find the instant at which x becomes 10 cm and y becomes 6 cm.
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We can write:
(i) 10 = x0 − 3t
(ii) 6 = y0 + 2t
• The two equations can be rearranged as shown below:
(iii) → [(i) × 2] → 20 = 2x0 − 6t
(iv) → [(ii) × 3] → 18 = 3y0 + 6t
(v) → [(iii) − (iv)] → 2 = 2x0 − 3y0 − 12t
⇒ 2x0 − 3y0 = 2 + 12t
4. Substituting the above result in (2), we get:
$\rm{\frac{dA}{dt}\,=\,2 + 12t - 12 t \,=\,2}$
• We can say that:
At the instant when x = 10 cm and y = 6 cm, the area is increasing at the rate of $\rm{2 ~ cm^2}$ per minute.
At any other instant, the rate will be different.
Solved example 22.5
The total cost C(x) in Rupees, associated with the production of x units of an item is given by:
C(x) = 0.005 x3 − 0.02x2 + 30x + 5000
Find the marginal cost when 3 units are produced. Marginal cost is the instantaneous rate of change of total cost at any level of output.
Solution:
Let us write some basic details about marginal cost. It can be written in 4 steps:
1. Consider the production of an item. By 'item', we mean products like LED bulb, bicycle, microwave oven etc.,
2. Take any one particular item. To produce 'x' no. of that item, the factory has to spend a certain amount of money.
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This amount (in Rupees) is given by:
C(x) = 0.005 x3 − 0.02x2 + 30x + 5000
3. We see that, there are four terms. Why is there four terms?
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One term may be related to the cost of raw materials required for production.
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Another term may be related to the wages of workers involved in the production.
For example, if Rupees 30/- is the labor charge for one number, then 30x has to be spend for the production of x numbers. So this term is related to the labor department.
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Another term may be related to the cost of electricity, water supply etc.,
4. However, we need not worry about what each term represent. We are asked to find the marginal cost.
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Marginal cost is the instantaneous rate of change of cost w.r.t number of items produced.
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So marginal cost is $\rm{\frac{dC}{dx}}$.
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In the numerator, we have change of cost. In the denominator, we have change in number. But we need not calculate numerator and denominator separately. We know how to calculate $\rm{\frac{dC}{dx}}$ directly.
Now we can do the calculations:
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We have:
$\rm{\frac{dC}{dx}\,=\,\frac{d}{dx} \left[0.005 x^3 - 0.02 x^2 + 30 x + 5000 \right] \,=\, 0.015 x^2 - 0.04 x + 30}$
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So when 3 units are produced, we get:
$\rm{\left. \frac{dC}{dx} \right|_{x = 3}\,=\,0.015 (3)^2 - 0.04 (3) + 30 \,=\,0.135 - 0.12 + 30 = \text{Rs} \, 30.015 \, \text{per number}}$
• We can say that:
At the instant when number of items produced is 3, the marginal cost is Rs 30.015.
At any other instant, the marginal cost will be different.
Solved example 22.6
The total revenue R(x) in Rupees, received from the sale of x units of a product is given by:
R(x) = 3 x2 + 36x + 5
Find
the marginal revenue when 5 units are sold. Marginal revenue is the
instantaneous rate of change of total revenue at any level of sale.
Solution:
Let us write some basic details about marginal revenue. It can be written in steps:
1. Consider the sale of a product. By 'product', we mean products like LED bulb, bicycle, microwave oven etc.,
2. Take any one particular product. When 'x' no. of that product is sold, the factory gets a certain amount of money.
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This amount (in Rupees) is given by:
R(x) = 3 x2 + 36x + 5
3. We see that, there are three terms. Why is there three terms?
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One term may be the actual amount received from the customer.
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Another term may be related to the amount received as commission.
4. However, we need not worry about what each term represent. We are asked to find the marginal revenue.
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Marginal revenue is the instantaneous rate of change of revenue w.r.t number of items sold.
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So marginal revenue is $\rm{\frac{dR}{dx}}$.
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In the numerator, we have change of revenue. In the denominator, we have
change in number. But we need not calculate numerator and denominator
separately. We know how to calculate $\rm{\frac{dR}{dx}}$ directly.
Now we can do the calculations:
• We have:
$\rm{\frac{dR}{dx}\,=\,\frac{d}{dx} \left[3 x^2 + 36 x + 5 \right] \,=\, 6 x + 36}$
• So when 5 units are sold, we get:
$\rm{\left. \frac{dR}{dx} \right|_{x = 5}\,=\,6(5) + 36 \,=\, = \text{Rs} \, 66 \, \text{per number}}$
• We can say that:
At the instant when number of items sold is 5, the marginal revenue is Rs 66.
At any other instant, the marginal revenue will be different.
The link below gives a few more solved examples:
In the next section, we will see increasing and decreasing functions.
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