23.1 - Geometrical Interpretation of Indefinite Integrals

In the previous section, we saw some basic details about integrals. In this section, we will see the geometrical interpretation of indefinite integrals.

The geometrical interpretation can be explained with the help of an example. It can be written in 5 steps:
1. Consider the function f(x) = 2x.
• For this function, the general form of the integral is: $\int{f(x) \, dx}~=~x^2\,+\,C$
2. The constant of integration C, can take infinite values.
• So there are infinite number of integrals. Every one of those integrals will have the same derivative.
• So every one of those integrals belong to a family of integrals.
• By assigning different values for C, we get the different members of the family.
• This family, with all the members, is called the indefinite integral.
3. For our present example, each member of the indefinite integral is a unique parabola.
• When C = 0,
   ♦ the parabola is: y = x²
   ♦ vertex of the parabola is at the origin.
• When C = 1,
   ♦ the parabola is: y = x² + 1
   ♦ vertex of the parabola is on the y-axis, one unit above the origin. 
• When C = 2,
   ♦ the parabola is: y = x² + 2
   ♦ vertex of the parabola is on the y-axis, two units above the origin. 
• When C = −1,
   ♦ the parabola is: y = x² − 1
   ♦ vertex of the parabola is on the y-axis, one unit below the origin. 
• When C = −2,
   ♦ the parabola is: y = x² − 2
   ♦ vertex of the parabola is on the y-axis, two units below the origin.
• So for each +ve value of C, the corresponding parabola has it's vertex on the +ve side of the y-axis.
• And for each −ve value of C, the corresponding parabola has it's vertex on the −ve side of the y-axis.
• This is shown in fig.23.1 below:

Fig.23.1

4. Mark all points where x-coordinate is 'a'. We will be able to draw a vertical line connecting all those points. Equation of that vertical line will be x = a.
• This vertical line will intersect all parabolas at unique points. At all those intersecting points, the slope of the parabola will be '2a'.
• If we draw tangents at those intersecting points, all those tangents will be parallel. Those parallel tangents are shown in magenta color in the fig.23.1 above. Note that in the fig., 'a' is assumed to be equal to '1.5'.
• This result is valid irrespective of whether 'a' is +ve or −ve.
5. We can pick any parabola from the family and move it up or down along the y-axis. It will be an exact fit over all other parabolas. This is shown in the animation below:

• In the animation, we pick the violet parabola from the previous fig.23.1, and move it vertically upwards and downwards. It is able to make an exact fit over each of the other parabolas.  

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Now we will see some properties of indefinite integrals.
Property I:
This can be written in 4 steps:
1. Let:
   ♦ F be the  anti derivative of f
   ♦ G be the  anti derivative g
• Then we can write:
$\frac{d}{dx}[F(x)\,+\,G(x)] ~=~\frac{d}{dx}[F(x)] ~+~ \frac{d}{dx}[G(x)] ~=~ f(x) \,+\, g(x)$

2. Based on this result, we can write:
[F(x) + G(x)] is the anti derivative of [f(x) + g(x)]
• That is:
$\int{[f(x) \,+\, g(x)]} ~=~ [F(x)\,+\,G(x)] \,+\,C$

4. So we can write Property I:
Antiderivative of a sum is same as the sum of the individual anti derivatives.

5. In the same way, we can prove that:
$\int{[f(x) \,-\, g(x)]} ~=~ [F(x)\,-\,G(x)] \,+\,C$

Property II:
This can be written in 6 steps:
1. Let:
   ♦ F be the  anti derivative of f
   ♦ k be any real number.
• Then we can write:
$\frac{d}{dx}[k F(x)] ~=~k \frac{d}{dx}[F(x)] ~=~ k f(x)$

2. Based on this result, we can write:
[k F(x)] is the anti derivative of [k f(x)]
• That is:
$\int{[k f(x)]} ~=~ k F(x) \,+\,C$

4. So we can write Property II:
Anti derivative of k times a function is same as k times the anti derivative of that function.

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Integration by method of inspection

This can be written in 3 steps:
1. Suppose that, we are asked to find the anti derivative of a given function f. That is., we are asked to find $\int{f(x) dx}$
• Then we can search for a function F such that, f is the derivative of F.
2. After finding F, we can straight away write:
$\int{f(x) dx}~=~F(x) ~+~ C$
3. This method of integration, is known as the integration by the method of inspection.

• An example can be written in 3 steps:
(i) Suppose that we are given f(x) = x
(ii) We search for F and find that, f(x) = x, is the derivative of $F(x)\,=\,\frac{x^2}{2}$
(iii) So we can write:
$\int{x \, dx}~=~\frac{x^2}{2}\,+\,C$ is the general form of the anti derivative of x.
• Or we can write:
$\int{x \, dx}~=~\frac{x^2}{2}$ is an anti derivative of x.

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Let us see some solved examples:

Solved example 23.1
Write an anti derivative for each of the following functions using the method of inspection:
(i) cos 2x    (ii) 3x² + 4x³    (iii) $\frac{1}{x}, x \, \ne \,0$
Solution:
Part (i):
1. We are given f(x) = cos 2x
2. We search for F and find that:
$\frac{d}{dx}\left[\frac{\sin 2x}{2} \right]~=~\cos 2x$
• That means:
f(x) = cos 2x, is the derivative of $F(x)\,=\,\frac{\sin 2x}{2}$
3. So we can write:
$\int{\cos 2x \, dx}~=~\frac{\sin 2x}{2}\,+\,C$ is the general form of the anti derivative of cos 2x.
• Or we can write:
$\int{\cos 2x \, dx}~=~\frac{\sin 2x}{2}$ is an anti derivative of cos 2x.

Part (ii):
1. We are given f(x) = 3x² + 4x³
2. We want $\int{\left[3x^2 \,+\, 4x^3 \right] dx}$
• By applying property I, this can be written as:
we want $\int{3x^2 dx} \,+\, \int{4x^3 dx}$
• We search for F and find that:
    ♦ 3x², is the derivative of x³
    ♦ 4x³, is the derivative of x⁴
3. So we can write:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[3x^2 \,+\, 4x^3 \right] dx}}    & {~=~}    &{\int{3x^2 dx} \,+\, \int{4x^3 dx}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{(x^3 \,+\, C_1) ~+~ (x^4 \,+\, C_2)  }    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{x^3 \,+\, x^4 \,+\, C}    \\
\end{array}$

• This is the general form of the required anti derivative.
◼ Remarks:
3 (magenta color): Addition of any two real numbers C₁ and C₂ will give another real number which can be denoted in general as C.

• Or we can write:
$\int{\left[3x^2 \,+\, 4x^3 \right] dx}~=~x^3 \,+\, x^4$ is an anti derivative of [3x² + 4x³].

Part (iii):
1. We are given $f(x) ~=~\frac{1}{x}, x \, \ne \,0$
2. We search for F and find that:
$\frac{d}{dx}\left[\log |x| \right]~=~\frac{1}{x}$
• That means:
$f(x) ~=~\frac{1}{x}, x \, \ne \,0$, is the derivative of $F(x)\,=\,\log |x|$
• We take the absolute value of x because, logarithm of −ve numbers are not defined.
3. So we can write:
$\int{\frac{1}{x} \, dx}~=~\log |x| \,+\,C$ is the general form of the anti derivative of $f(x) ~=~\frac{1}{x}, x \, \ne \,0$.
• Or we can write:
$\int{\frac{1}{x} \, dx}~=~\log |x|$ is an anti derivative of $f(x) ~=~\frac{1}{x}, x \, \ne \,0$.

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In the next section, we will see a few more solved examples.

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Chapter 23 - Integrals

In the previous section, we completed a discussion on applications of derivatives. In this chapter, we will see Integrals.

• We began our discussion on derivatives in section 13.10.
• There we saw the equation for finding the distance traveled by a freely falling body as: s = 4.9t².
• Since the distance s depends on time t, we can say that, s is a function of t. We write this as: s = f (t) = 4.9t²
• Derivative of this function is given by: ${\frac{ds}{dt}\,=\,f'(t) \,=\,9.8t}$
• This derivative can be used to find the instantaneous velocity at any instant t. For example, at the instant when the stop-watch shows 8 s, the velocity will be: f '(8) = 9.8(8) = 78.4 m/s.
• Now let us think in reverse:
If we are given the derivative, can we find the original function?
This chapter tries to answer the above question.

The process of integration
• Suppose that, we are given a derivative f '(x).
   ♦ The process of finding the original function f(x) is known as integration.
   ♦ This process is also known as anti differentiation.
• In fact, integration is the inverse process of differentiation.
• The original function that we obtain by integration is called primitive.

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Let us see some basic details about integration. It can be written in 5 steps:
1. First we will see three examples:
Example (i): We know that ${\frac{d}{dx}(\sin x)\,=\,\cos x}$
• That is., cos x is the derivative of sin x.
• This is same as: sin x is an anti derivative of cos x
   ♦ An anti derivative is also called an integral.       
Example (ii): We know that ${\frac{d}{dx}\left(\frac{x^3}{3} \right)\,=\,x^2}$
• That is., x² is the derivative of ${\frac{x^3}{3}}$.
• This is same as: ${\frac{x^3}{3}}$ is an integral of x².
Example (iii): We know that ${\frac{d}{dx}\left(e^x \right)\,=\,e^x}$
• That is., e^x is the derivative of e^x.
• This is same as: e^x is an integral of e^x.

2. Recall that, the derivative of a constant function is zero.
• Now consider the example (i) that we saw above.
The original function may be the sum of sin x and a constant function C (where C is any real number). Even then, we will get the same derivative:
${\frac{d}{dx}(\sin x\,+\,C)\,=\,\cos x \,+\, 0 \,=\, \cos x}$
So the integral is: sin x + C
• Similarly, consider the example (ii).
The original function may be the sum of ${\frac{x^3}{3}}$ and a constant function C (where C is any real number). Even then, we will get the same derivative:
${\frac{d}{dx}\left(\frac{x^3}{3} \,+\, C \right)\,=\,x^2 \,+\, 0 \,=\, x^2}$
So the integral is ${\frac{x^3}{3} \,+\, C}$
• Similarly, consider the example (iii).
The original function may be the sum of ${e^x}$ and a constant function C (where C is any real number). Even then, we will get the same derivative:
${\frac{d}{dx}(e^x \,+\, C )\,=\,e^x \,+\, 0 \,=\, e^x}$
So the integral is ${e^x \,+\, C}$

3. The constant C can be chosen arbitrarily from the set of real numbers. That means, C can be any random real number.
• So there are infinite possibilities for the integral. That means, there will be infinite integrals for a given derivative.
• For each of those infinite integrals, the derivative will be the same.
   ♦ The constant C is called arbitrary constant.
   ♦ The constant C is also called constant of integration.
• By varying C, we can get infinite anti derivatives (integrals) for any given derivative.

4. Since there are infinite integrals with the same derivative, we will write a general form. It can be explained in 4 steps:
(i) Suppose that, we are given a derivative.
• Recall that, derivative is also a function. So we will denote the given derivative as f(x).
(ii) Let the integral obtained by the integration of f(x), be F(x).
• Then we can write: $\frac{d}{dx} \left[F(x) \,+\, C \right]\,=\,f(x)$
   ♦ Where C is the constant of integration.
(iii) From the above expression, it is clear that:
By the integration of f(x), we will get F(x) + C
• This [F(x) + C] is a function.
   ♦ F(x) is unique.
   ♦ C can have infinite values.
(iv) Since C can have infinite values, there are infinite functions of the form [F(x) + C]
• We can write a set of all those functions as: {F+C, C ∈ R}
• This set is the general form of all integrals of f.

5. Now we will see an important property of integrals. It can be written in 6 step:
(i) Suppose that, we are given a function f.
• Pick any two integrals from among the infinite possible integrals. Let the two picked integrals be:
   ♦ F(x) + C₁
   ♦ F(x) + C₂
(ii) So we have two functions: F(x) + C₁ and F(x) + C₂
• Let us name them as g and h. So we have:
   ♦ g(x) = F(x) + C₁
   ♦ h(x) = F(x) + C₂
(iii) Now we can write the difference of g and h. We get:
g(x) − h(x) = [F(x) + C₁] − [F(x) + C₂] = [C₁ − C₂]
(iv) But [C₁ − C₂] is a constant. Let us call it C₃.
• So we can write: g(x) − h(x) = C₃
(v) Let us differentiate [g(x) − h(x)]. We get:
$\frac{d}{dx}[g(x) - h(x)]\,=\,\frac{d}{dx}[C_3]\,=\,0$
(vi) From (ii), we know that, g and h have the same derivative.
• So we can write:
If g and h are any two functions which have the same derivative, then the derivative of the "difference of g and h" will be zero.

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• We have seen the basics of integration. In short, we can write:
We are given the derivative f(x). The process of finding the corresponding [F(x) + C] is called integration.

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• We use the symbol $\int{f(x) \, dx}$ to denote the process of integration.
• When we see this symbol, we say:
Integration of f with respect to x.
• So we can write: $\int{f(x) \, dx}~=~F(x) \,+\,C$

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Following table shows the meanings of various symbols/terms/phrases related to integration.

Table 23.1

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• When we learned about derivatives, we saw many useful formulas. For example: $\frac{d}{dx} (\tan x) \,=\, \sec^2 x$.
• This formula can be used to write the corresponding formula for integral. We get: $\int{\sec^2 x \, dx}\,=\,\tan x \,+\, C$
• In this way, using the already known derivative formulas, we can make a list of integral formulas. A small list is given below:

$\begin{array}{ll} {~\color{magenta}    {}    }    &{{}}    &{\text{Derivatives}}    & {{}}    &{\text{Integrals}}    \\
{~\color{}    (i)    }    &{{}}    &{\frac{d}{dx}\left(\frac{x^{n+1}}{n+1} \right)~=~x^n}    & {\implies}    &{\int{x^n \, dx}~=~ \frac{x^{n+1}}{n+1}  \,+\, C}    \\
{~\color{}    {}    }    &{{}}    &{{}}    & {{}}    &{{}}    \\
{~\color{}    (ii)    }    &{{}}    &{\frac{d}{dx}\left(x \right)~=~1}    & {\implies}    &{\int{(1) \, dx}~=~ x \,+\, C}    \\
{~\color{magenta}    {}    }    &{{}}    &{{}}    & {{}}    &{{}}    \\
{~\color{}    (iii)    }    &{{}}    &{\frac{d}{dx}\left(\sin x \right)~=~\cos x}    & {\implies}    &{\int{\cos x \, dx}~=~ \sin x \,+\, C}    \\
{~\color{magenta}    {}    }    &{{}}    &{{}}    & {{}}    &{{}}    \\
{~\color{}    (iv)    }    &{{}}    &{\frac{d}{dx}\left(\cos^{-1} x \right)~=~ \frac{-1}{\sqrt{1 - x^2}}}    & {\implies}    &{\int{\frac{-1}{\sqrt{1 - x^2}} \, dx}~=~ \cos^{-1} x \,+\, C}    \\
{~\color{magenta}    {}    }    &{{}}    &{{}}    & {{}}    &{{}}    \\
{~\color{}    (v)    }    &{{}}    &{\frac{d}{dx}\left(\log |x| \right)~=~\frac{1}{x}}    & {\implies}    &{\int{\frac{1}{x} \, dx}~=~ \log |x| \,+\, C}    \\
{~\color{}    {}    }    &{{}}    &{{}}    & {{}}    &{{}}    \\
\end{array}$

• The reader may add as many items as possible, to this list. We will be using the items in the list to solve complicated problems.

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In the next section, we will see geometrical interpretation of indefinite integral.

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22.22 - Miscellaneous Examples on Applications of Derivatives - Part 4

In the previous section, we saw some miscellaneous examples. In this section, we will see a few more miscellaneous examples.

Solved example 22.82
A circular disc of radius 3 cm is being heated. Due to expansion, it's radius increases at the rate of 0.05 cm/s. Find the rate at which its area is increasing when radius is 3.2 cm.
Solution:
1. We have the formula for area of a circle: A = πr².
2. The radius increases at the rate of 0.05 cm/s.
• So radius at time t will be 0.05t
4. So area A at time t = π(0.05t)² = 0.0025πt² cm²
• Now the rate of change of area w.r.t radius can be obtained as:
dA/dt = 0.0025π(2t) = 0.005 πt cm²/s
• This result can be used to find the rate of change of area at any instant.
5. We want the rate when radius is 3.2 cm. So we want the instant at which the radius is 3.2 cm. For that, we can use the result in (2). We get:
3.2 = 0.05t
⇒ t = 64
6. So from (4) we get:
(dA/dt)_t=64 = 0.005π(64) = 0.320π cm²/s

Solved example 22.83
An open topped box is to be constructed by removing equal squares from each corner of a 3 metre by 8 metre rectangular sheet of aluminium and folding up the sides. Find the volume of the largest such box.
Solution:
Step I: Writing the problem as a function
1. In fig.22.77 below, the original rectangular sheet is shown in yellow color. Red squares are cut off from the corners.

Fig.22.77

• So when the flaps are folded up, we will get a box with:
   ♦ base a rectangle (8 − 2x) × (3 − 2x) m²
   ♦ height x m
• Then the volume of the box = x(8 − 2x)(3 − 2x) m³

2. It is clear that, V depends on x. That means, V is a function of x. So we can write:
V = f(x) = x(8 − 2x)(3 − 2x) m³

3. We want V to be as large as possible. That means, we want the largest of all local maxima of f. That means, we need to find all local maxima and compare them.

Step II: Finding the critical points
1. Write the first two derivatives:
• f '(x) =12x² − 44x + 24

• f ''(x) = 24x − 44

2. Equate the first derivative to zero and solve for x:
12x² − 44x + 24 = 0
⇒ x = 2/3, x = 3 

3. So the only two points in category I are: x = 2/3 and x = 3
4. We obtained f '(x) = 12x² − 44x + 24
• This function is defined for all real numbers. That means, there is no input x at which f '(x) is not defined.
• Therefore, there will be no points in category II.
5. So the only two critical points are: x = 2/3 and x = 3

Step III: Find f ''(x) at the critical points
1. f ''(2/3) = 24(2/3) − 44 = −28
2. f ''(3) =  24(3) − 44 = 28

Step IV: Applying the second derivative test
1. Since f ''(2/3) is −ve, there is a local maximum at x = 2/3
2. Since f ''(3) is +ve, there is a local minimum at x = 3

Step V: Finding the actual maximum value:
• The local maximum value at x = 2/3 is given by:
f(2/3) = (2/3)[8 − 2(2/3)][3 − 2(2/3)] = 200/27 m³.

Solved example 22.84
Manufacturer can sell x items at a price of rupees $\rm{5 - \frac{x}{100}}$ each. The cost price of x items is Rs $\rm{\frac{x}{5} + 500}$. Find the number of items he should sell to earn maximum profit.
Solution:
Step I: Writing the problem as a function
1. When the manufacturer sells x items, he will get an income of: Rs $\rm{x\left(5 - \frac{x}{100} \right)}$
2. To produce those x items, he would have spent:
Rs $\rm{\frac{x}{5} + 500}$
3. So the profit P can be obtained as:
$\rm{P\,=\, x\left(5 - \frac{x}{100} \right)~-~\frac{x}{5} + 500}$
4. It is clear that, P depends on x. That means, P is a function of x. So we can write:
$\rm{P\,=\, f(x)\,=\,x\left(5 - \frac{x}{100} \right)~-~\frac{x}{5} + 500}$
5. We want P to be as large as possible. That means, we want the largest of all local maxima of f. That means, we need to find all local maxima and compare them.

Step II: Finding the critical points
1. Write the first two derivatives:
• f '(x) = $\rm{\frac{-x}{50} + \frac{24}{5}}$

• f ''(x) = $\rm{\frac{-1}{50}}$

2. Equate the first derivative to zero and solve for x:
$\rm{\frac{-x}{50} + \frac{25}{4} \,=\, 0}$
⇒ x = 240

3. So the only one point in category I is: x = 240
4. We obtained f '(x) = $\rm{\frac{-x}{50} + \frac{24}{5}}$
• This function is defined for all real numbers. That means, there is no input x at which f '(x) is not defined.
• Therefore, there will be no points in category II.
5. So the only one critical point is: x = 240

Step III: Find f ''(x) at the critical points
• f ''(x) is a constant: $\rm{\frac{-1}{50}}$

Step IV: Applying the second derivative test
• Since f ''(240) is −ve, there is a local maximum at x = 240

Step V: Final result:
• For maximum profit, the manufacturer must sell 240 items.

Solved example 22.85
Find the maximum area of an isosceles triangle inscribed in the ellipse $\rm{\frac{x^2}{a^2}\,+\,\frac{y^2}{b^2}\,=\,1}$ with it's vertex at one end of the major axis.
Solution:
Step I: Writing the problem as a function
1. The given ellipse is: $\rm{\frac{x^2}{a^2}\,+\,\frac{y^2}{b^2}\,=\,1}$
• So it is clear that:
    ♦ Major axis lies along the x-axis.
    ♦ Minor axis lies along the y-axis.
    ♦ Center of ellipse is at the origin O.
2. Thus we get fig.22.78 below:

Fig.22.78

• The vertex A of the isosceles triangle is at one end of the major axis.
3. Let the base of the isosceles triangle intersect the x-axis at D(h,0).
• Then the y-coordinates of B and C can be obtained as shown below:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{x^2}{a^2}\,+\,\frac{y^2}{b^2}}    & {~=~}    &{1}    \\
{~\color{magenta}    2    }    &{\implies}    &{\frac{h^2}{a^2}\,+\,\frac{y^2}{b^2}}    & {~=~}    &{1}    \\
{~\color{magenta}    3    }    &{\implies}    &{\frac{y^2}{b^2}}    & {~=~}    &{1~-~\frac{h^2}{a^2}}    \\
{~\color{magenta}    4    }    &{\implies}    &{y^2}    & {~=~}    &{b^2 \left(1~-~\frac{h^2}{a^2} \right)}    \\
{~\color{magenta}    5    }    &{\implies}    &{y}    & {~=~}    &{\pm \sqrt{b^2 \left(1~-~\frac{h^2}{a^2} \right)}}    \\
\end{array}$

• So the y-coordinate of B is $\sqrt{b^2 \left(1~-~\frac{h^2}{a^2} \right)}$
• And the y-coordinate of C is $-\sqrt{b^2 \left(1~-~\frac{h^2}{a^2} \right)}$
• Then the distance BC will be $2 \sqrt{b^2 \left(1~-~\frac{h^2}{a^2} \right)}$
4. Next we want the coordinates of A. It can be calculated as shown below:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{x^2}{a^2}\,+\,\frac{y^2}{b^2}}    & {~=~}    &{1}    \\
{~\color{magenta}    2    }    &{\implies}    &{\frac{x^2}{a^2}\,+\,\frac{0^2}{b^2}}    & {~=~}    &{1}    \\
{~\color{magenta}    3    }    &{\implies}    &{\frac{x^2}{a^2}}    & {~=~}    &{1}    \\
{~\color{magenta}    4    }    &{\implies}    &{x^2}    & {~=~}    &{a^2}    \\
{~\color{magenta}    5    }    &{\implies}    &{x}    & {~=~}    &{\pm a}    \\
\end{array}$

• So the coordinates of A are: (a,0).
• Then the altitude AD = (h+a)
5. Now we can write the expression for the area of triangle ABC:
Area = 1/2 × Base × Altitude = 1/2 × BC × AD
= $\frac{1}{2} × 2 \sqrt{b^2 \left(1~-~\frac{h^2}{a^2} \right)} × (h+a)$
• This can be simplified as:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\text{Area}}    & {~=~}    &{\frac{1}{2} × 2 \sqrt{b^2 \left(1~-~\frac{h^2}{a^2} \right)} × (h+a)}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{b \sqrt{\left(1~-~\frac{h^2}{a^2} \right)} × (h+a)}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{b \sqrt{\left(\frac{a^2 – h^2}{a^2} \right)} × (h+a)}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{b}{a} \sqrt{\left(a^2 – h^2 \right)} × (h+a)}    \\
\end{array}$

6. It is clear that, Area A depends on h. That means, A is a function of h. So we can write:
$\rm{A\,=\, f(x)\,=\,\frac{b}{a} \sqrt{\left(a^2 – h^2 \right)} × (h+a)}$
5. We want A to be as large as possible. That means, we want the largest of all local maxima of f. That means, we need to find all local maxima and compare them.

Step II: Finding the critical points
1. Write the first two derivatives:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f(h)}    & {~=~}    &{\frac{b}{a} \sqrt{\left(a^2 – h^2 \right)} × (h+a)}    \\
{~\color{magenta}    2    }    &{\implies}    &{f'(h)}    & {~=~}    &{\frac{b}{a} \left[\sqrt{\left(a^2 – h^2 \right)} × (1) ~+~(h+a) (1/2) \left(a^2 – h^2 \right)^{-1/2} (-2h) \right]}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{b}{a} \left[\sqrt{\left(a^2 – h^2 \right)}  ~-~ \frac{h(h+a)}{\sqrt{\left(a^2 – h^2 \right)}} \right]}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{b}{a} \left[\frac{a^2 – h^2 ~-~h(h+a)}{\sqrt{\left(a^2 – h^2 \right)}} \right]}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\frac{b}{a} \left[\frac{a^2 – h^2 – h^2 – ha}{\sqrt{\left(a^2 – h^2 \right)}} \right]}    \\
{~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{\frac{b}{a} \left[\frac{a^2 – 2 h^2  – ha}{\sqrt{\left(a^2 – h^2 \right)}} \right]}    \\
\end{array}$                           

• In this problem, calculation of f ''(h) will be a lengthy process. So we will use the first derivative test

2. Equate the first derivative to zero and solve for h:
$\rm{a^2 – 2 h^2  – ha \,=\, 0}$
⇒ $\rm{2 h^2 + ha - a^2 \,=\, 0}$
• Using quadratic formula, we get:
h = a/2 and h = −a
• h cannot be −a. Because, then in fig.22.78 above, D will coincide with the left end of the major axis. Parts of the triangle will fall outside the ellipse.

3. So the only one point in category I is: x = a/2
4. We obtained f '(h) = $\frac{b}{a} \left[\frac{a^2 – 2 h^2  – ha}{\sqrt{\left(a^2 – h^2 \right)}} \right]$
• This function is not defined when h = a. But when h = a, the function itself becomes invalid because there will be no inscribed triangle.
• Therefore, there will be no points in category II.
5. So the only one critical point is: h = a/2
6. The domain is (−a,a).
• Based on the critical point, the domain can be divided into two intervals: (−a, a/2) and (a/2, a)

Step III: Analyzing the signs of f '(x) in the two intervals
1. Consider the first interval. Zero is a convenient point.
• We get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f'(h)}    & {~=~}    &{\frac{b}{a} \left[\frac{a^2 – 2 h^2  – ha}{\sqrt{\left(a^2 – h^2 \right)}} \right]}    \\
{~\color{magenta}    2    }    &{\implies}    &{f'(0)}    & {~=~}    &{\frac{b}{a} \left[\frac{a^2 – 2 (0)^2  – (0)a}{\sqrt{\left(a^2 – (0)^2 \right)}} \right]}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{b}{a} \left[\frac{a^2}{\sqrt{\left(a^2 \right)}} \right]}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{b}    \\
\end{array}$

• So in the first interval. f '(x) is +ve.

2. Consider the second interval. $\frac{3a}{4}$  is a convenient point.
• We get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f'(h)}    & {~=~}    &{\frac{b}{a} \left[\frac{a^2 – 2 h^2  – ha}{\sqrt{\left(a^2 – h^2 \right)}} \right]}    \\
{~\color{magenta}    2    }    &{\implies}    &{f'(3a/4)}    & {~=~}    &{\frac{b}{a} \left[\frac{a^2 – 2 (3a/4)^2  – (3a/4)a}{\sqrt{\left(a^2 – (3a/4)^2 \right)}} \right]}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{b}{a} \left[\frac{a^2 – 2 \left(\frac{9 a^2}{16} \right)  – \frac{3 a^2}{4}}{\sqrt{\left(a^2 – \frac{9 a^2}{16} \right)}} \right]}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{b}{a} \left[\frac{\frac{16 a^2 - 18 a^2 – 12 a^2}{16}}{\sqrt{\left(\frac{16 a^2 – 9 a^2}{16} \right)}} \right]}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\frac{b}{a} \left[\frac{\frac{16 a^2 - 30 a^2}{16}}{\sqrt{\left(\frac{7 a^2}{16} \right)}} \right]}    \\
{~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{\frac{b}{a} \left[\frac{\frac{- 14 a^2}{16}}{\sqrt{\left(\frac{7 a^2}{16} \right)}} \right]}    \\
\end{array}$

• So in the second interval. f '(x) is −ve.

Step IV: Applying the first derivative test
• Since f '(x) changes sign from +ve to −ve at the critical point, we have a maximum at that critical point.

Step V: Final result:
• For maximum area, h = a/2
So from step I, we get:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\text{Area}}    & {~=~}    &{\frac{1}{2} × 2 \sqrt{b^2 \left(1~-~\frac{h^2}{a^2} \right)} × (h+a)}    \\
{~\color{magenta}    2    }    &{\implies}    &{\text{Area (max)}}    & {~=~}    &{\sqrt{b^2 \left(1~-~\frac{(a/2)^2}{a^2} \right)} × ((a/2)+a)}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\sqrt{b^2 \left(1~-~\frac{a^2 / 4}{a^2} \right)} × (3a/2)}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\sqrt{3 b^2 / 4} × (3a/2)}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\frac{\sqrt{3} b}{2} × \frac{3a}{2}}    \\
{~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{\frac{3 \sqrt{3} \,ab}{4}}    \\
\end{array}$

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The link below gives a few more miscellaneous examples:

Miscellaneous Exercise

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We have completed this chapter on applications of derivatives. In the next chapter, we will see Integrals.

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