Wednesday, October 16, 2024

22.6 - Linear Approximation

In the previous section, we completed a discussion on Tangents and Normals. In this section, we will see Approximations.

Some basic details can be written in 12 steps:
1. Fig.22.21 below shows the graph of f(x) = x3 − 6x + 2
It is drawn in red color.

Fig.22.21


2. Consider point A where x = 2
We get: f(2) = (2)3 − 6(2) + 2 = −2
3. The green line is the tangent at x = 2
• Let us find the equation of this tangent.
f'(x) = 3x2` − 6
⇒ f'(2) = 3(2)2` − 6 = 6
y − (−2) = 6(x − 2)
⇒ y + 2 = 6 x − 12
⇒ y = 6x − 14
4. This tangent can be considered as a function of x. We will name this function as L. So we can write:
L(x) = 6x − 14
5. Let us calculate the value of L at x = 2
L(2) = 6(2) − 14 = −2
6. From (2) and (5), we can write: f(2) = L(2)
• In general, we can write: f(a) = L(a)
This is possible because, the tangent L is drawn at x = a
7. Consider another point B where x = 1.9.
    ♦ At A, x = 2
    ♦ At B, x = 1.9
• So we can say that, B is close to A
    ♦ f(1.9) = (1.9)3 − 6(1.9) + 2 = −2.541
    ♦ L(1.9) = 6(1.9) − 14 = −2.6
• We see that, f(1.9) is approximately equal to L(1.9)
• When we calculate L(1.9), we get the coordinates of a new point as (1.9, − 2.6). We will name this point as B'.
In the graph, we see that, B is very close to B'.
8. From the above discussion, we can write 3 points:
(i) We wanted the value of f at B. For that, we used the tangent at A
(ii) If we directly use f, we get: f(1.9) = −2.541
(iii) If we use the tangent at A, we get: L(1.9) = −2.6
(iv) We can write: f(1.9) ≈ L(1.9)
9. We accept the fact that, there is considerable difference between −2.541 and −2.6
• This difference is due to the fact that, B is not very close to A.
    ♦ x value at B is 1.9
    ♦ x value at A is 2
• If B is very close to A, we will get better approximations. Let us try such values for B:
(i) At B, if x = 1.95, we get:
f(1.95) = −2.2851 and L(1.95) = −2.3
(ii) At B, if x = 1.98, we get:
f(1.98) = −2.1176 and L(1.98) = −2.12
(iii) At B, if x = 1.99, we get:
f(1.99) = −2.059401 and L(1.99) = −2.06
(iv) At B, if x = 1.999, we get:
f(1.999) = −2.005994001 and L(1.999) = −2.006
(iv) At B, if x = 1.9999, we get:
f(1.9999) = −2.000599940001 and L(1.999) = −2.0006
10. It is clear that, if B is very close to A, then the tangent at A can be used to find the approximate value of f at B.
• The method can be summarized in 6 steps:
(i) We want the approximate value of f at B. Assume that, at B, x = b.
• So we want the approximate value of f(b).
(ii) For that, we choose a convenient point A, which is very close to B.
• Assume that, at A, x = a
(iii) We write the function L at A.
(iv) Then the approximate value of f(b) will be L(b)
(v) In the function f, the power of x may be one, two or above. But in the function L, the power of x will be always one. So it is easier to find L(b) than to find f(b).
(vi) The function L is known by three different names. We can use any on of them
   ♦ L is called linear approximation of f at x = a.
   ♦ L is called tangent line approximation of f at x = a.
   ♦ L is called linearization of f at x = a.
11. This method will work also when B (very close to A), is to the right of A. Let us see such points.
(i) At B, if x = 2.05, we get:
f(2.05) = −1.6848 and L(2.05) = −1.7
(ii) At B, if x = 2.02, we get:
f(2.02) = −1.877592 and L(2.02) = −1.88
(iii) At B, if x = 2.01, we get:
f(2.01) = −-1.939399 and L(2.01) = −1.94
(iv) At B, if x = 2.001, we get:
f(2.001) = −-1.993993999 and L(2.001) = −1.994
(iv) At B, if x = 2.0001, we get:
f(2.0001) = −1.999399939999 and L(2.0001) = −1.9994
12. This method will work only if B is very close to A. In the fig.22.21 above, a point C is marked. This point C is not close to A.
    ♦ x value at C is 2.4
    ♦ x value at A is 2
• We see that:
    ♦ f(2.4) is 1.424
    ♦ L(2.4) = 0.4
• There is a very large difference between 1.424 and 0.4
• This is because, there is a large difference between 2 and 2.4.


Let us see some solved examples:

Solved example 22.21
Write the linear approximation of f(x) = √x and use it to estimate $\sqrt{36.6}$.
Solution:
1. Given $\rm{f(x) = \sqrt{x}}$
• We are asked to write the linear approximation of this function at an arbitrary point. Let us choose the arbitrary point x = a.
2. The slope at any point is $f'(x) ~=~\rm{\frac{1}{2} (x)^{-1/2}}$  
• So the slope at 'a' is $\rm{f'(a) ~=~\frac{1}{2} (a)^{-1/2}~=~\frac{1}{2 \sqrt{a}}}$
3. At the point where x = a, the y coordinate can be calculated as follows:
y = √x = √a
4. So the tangent at x = a can be obtained as:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y - \sqrt{a}}    & {~=~}    &{\frac{1}{2 \sqrt{a}} (x - a)}    \\
{~\color{magenta}    2    }    &{\implies}    &{y}    & {~=~}    &{\frac{1}{2 \sqrt{a}} (x - a) + \sqrt{a}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{x}{2 \sqrt{a}} ~-~\frac{a}{2 \sqrt{a}} ~+~ \sqrt{a}}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{x}{2 \sqrt{a}} ~-~\frac{\sqrt{a}}{2} ~+~ \sqrt{a}}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\frac{x}{2 \sqrt{a}} ~+~\frac{\sqrt{a}}{2}}    \\
\end{array}$
5. So the linear approximation of f at x = a is:
$L(x) = \frac{x}{2 \sqrt{a}} ~+~\frac{\sqrt{a}}{2}$
6. Using linear approximation, we can calculate f(b) because:
f(b) ≈ L(b) if b is close to a.
• In our present case, b = 36.6
• 'a' should be selected in such a way that:
   ♦ 'a' is a convenient number
   ♦ 'a' is close to 'b'.
• We can take a = 36
Square root of 36 is already known. So it is a convenient number. Also, 36 is close to 36.6
7. Based on the result in (5), we can write:
Linear approximation of f at x = 36 is:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{L(x)}    & {~=~}    &{\frac{x}{2 \sqrt{a}} ~+~\frac{\sqrt{a}}{2}}    \\
{~\color{magenta}    2    }    &{\implies}    &{L(x)_{a=36}}    & {~=~}    &{\frac{x}{2 \sqrt{36}} ~+~\frac{\sqrt{36}}{2}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{x}{2 (6)} ~+~\frac{6}{2}}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{x}{12} ~+~3}    \\
\end{array}$
8. Based on what we wrote in (6), we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f(b)}    & {~≈~}    &{L(b)}    \\
{~\color{magenta}    2    }    &{\implies}    &{f(36.6)}    & {~≈~}    &{L(36.6)}    \\
{~\color{magenta}    3    }    &{\implies}    &{\sqrt{36.6}}    & {~≈~}    &{\frac{36.6}{12} ~+~3}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~≈~}    &{3.05 ~+~ 3}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~≈~}    &{6.05}    \\
\end{array}$
9. It is interesting to note that, if we use a calculator or computer to calculate the square root of 36.6, we will get:
$\sqrt{36.6}~=~6.0497$
10. Fig.22.22 below shows the graph.
   ♦ f is drawn in red color
   ♦ L is drawn in green color

Fig.22.22

   ♦ A is a point on f
   ♦ B is a point on f
   ♦ B' is a point on L
• We want the y-coordinate at B. But, for ease of calculation, we use linear approximation and obtain the y-coordinate at B'.
• In the graph, B and B' are so close to each other that, B' is not distinctly visible from B.  

Solved example 22.22
Write the linear approximation of f(x) = √x and use it to estimate √(9.1).
Solution:
1. Given $\rm{f(x) = \sqrt{x}}$
• We are asked to write the linear approximation of this function at an arbitrary point. Let us choose the arbitrary point x = a.
2. The slope at any point is $f'(x) ~=~\rm{\frac{1}{2} (x)^{-1/2}}$  
• So the slope at 'a' is $\rm{f'(a) ~=~\frac{1}{2} (a)^{-1/2}~=~\frac{1}{2 \sqrt{a}}}$
3. At the point where x = a, the y coordinate can be calculated as follows:
y = √x = √a
4. So the tangent at x = a can be obtained as:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y - \sqrt{a}}    & {~=~}    &{\frac{1}{2 \sqrt{a}} (x - a)}    \\
{~\color{magenta}    2    }    &{\implies}    &{y}    & {~=~}    &{\frac{1}{2 \sqrt{a}} (x - a) + \sqrt{a}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{x}{2 \sqrt{a}} ~-~\frac{a}{2 \sqrt{a}} ~+~ \sqrt{a}}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{x}{2 \sqrt{a}} ~-~\frac{\sqrt{a}}{2} ~+~ \sqrt{a}}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\frac{x}{2 \sqrt{a}} ~+~\frac{\sqrt{a}}{2}}    \\
\end{array}$
5. So the linear approximation of f at x = a is:
$L(x) = \frac{x}{2 \sqrt{a}} ~+~\frac{\sqrt{a}}{2}$
6. Using linear approximation, we can calculate f(b) because:
f(b) ≈ L(b) if b is close to a.
• In our present case, b = 9.1
• 'a' should be selected in such a way that:
   ♦ 'a' is a convenient number
   ♦ 'a' is close to 'b'.
• We can take a = 9
Square root of 9 is already known. So it is a convenient number. Also, 9 is close to 9.1
7. Based on the result in (5), we can write:
Linear approximation of f at x = 9 is:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{L(x)}    & {~=~}    &{\frac{x}{2 \sqrt{a}} ~+~\frac{\sqrt{a}}{2}}    \\
{~\color{magenta}    2    }    &{\implies}    &{L(x)_{a=9}}    & {~=~}    &{\frac{x}{2 \sqrt{9}} ~+~\frac{\sqrt{9}}{2}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{x}{2 (3)} ~+~\frac{3}{2}}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{x}{6} ~+~\frac{3}{2}}    \\
\end{array}$
8. Based on what we wrote in (6), we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f(b)}    & {~≈~}    &{L(b)}    \\
{~\color{magenta}    2    }    &{\implies}    &{f(9.1)}    & {~≈~}    &{L(9.1)}    \\
{~\color{magenta}    3    }    &{\implies}    &{\sqrt{9.1}}    & {~≈~}    &{\frac{9.1}{6} ~+~\frac{3}{2}}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~≈~}    &{1.51667 ~+~ 1.5}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{3.01667}    \\
\end{array}$
9. It is interesting to note that, if we use a calculator or computer to calculate the square root of 9.1, we will get:
$\sqrt{9.1}~=~3.0166$

Solved example 22.23
Write the linear approximation of $\rm{f(x)\,=\,\sqrt[3]{x}}$ and use it to estimate $\rm{\sqrt[3]{8.1}}$.
Solution:
1. Given $\rm{f(x)\,=\,\sqrt[3]{x}}$
• We are asked to write the linear approximation of this function at an arbitrary point. Let us choose the arbitrary point x = a.
2. The slope at any point is $f'(x) ~=~\rm{\frac{1}{3} (x)^{-2/3}}$  
• So the slope at 'a' is $\rm{f'(a) ~=~\frac{1}{3} (a)^{-2/3}}$
3. At the point where x = a, the y coordinate can be calculated as follows:
$\rm{y\,=\,\sqrt[3]{a}\,=\,a^{1/3}}$
4. So the tangent at x = a can be obtained as:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y - a^{1/3}}    & {~=~}    &{\frac{1}{3} (a)^{-2/3} (x - a)}    \\
{~\color{magenta}    2    }    &{\implies}    &{y}    & {~=~}    &{\frac{1}{3} (a)^{-2/3} (x) ~-~ \frac{1}{3} (a)^{-2/3}(a)~+~a^{1/3}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{3} (a)^{-2/3} (x) ~-~ \frac{1}{3} a^{1/3}~+~a^{1/3}}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{3} (a)^{-2/3} (x) ~+~ \frac{2}{3} (a)^{1/3}}    \\
\end{array}$
5. So the linear approximation of f at x = a is:
$L(x) = \frac{1}{3} (a)^{-2/3} (x) ~+~ \frac{2}{3} (a)^{1/3}$
6. Using linear approximation, we can calculate f(b) because:
f(b) ≈ L(b) if b is close to a.
• In our present case, b = 8.1
• 'a' should be selected in such a way that:
   ♦ 'a' is a convenient number
   ♦ 'a' is close to 'b'.
• We can take a = 8
Cube root of 8 is already known. So it is a convenient number. Also, 8 is close to 8.1
7. Based on the result in (5), we can write:
Linear approximation of f at x = 8 is:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{L(x)}    & {~=~}    &{\frac{1}{3} (a)^{-2/3} (x) ~+~ \frac{2}{3} (a)^{1/3}}    \\
{~\color{magenta}    2    }    &{\implies}    &{L(x)_{a=8}}    & {~=~}    &{\frac{1}{3} (8)^{-2/3} (x) ~+~ \frac{2}{3} (8)^{1/3}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{3} (2)^{-2} (x) ~+~ \frac{2}{3} (2)}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{x}{12} ~+~\frac{4}{3}}    \\
\end{array}$
8. Based on what we wrote in (6), we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f(b)}    & {~≈~}    &{L(b)}    \\
{~\color{magenta}    2    }    &{\implies}    &{f(8.1)}    & {~≈~}    &{L(8.1)}    \\
{~\color{magenta}    3    }    &{\implies}    &{\sqrt[3]{8.1}}    & {~≈~}    &{\frac{8.1}{12} ~+~\frac{4}{3}}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~≈~}    &{0.675 ~+~ 1.33333}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~≈~}    &{2.00833}    \\
\end{array}$                           
9. It is interesting to note that, if we use a calculator or computer to calculate the cube root of 8.1, we will get:
$\sqrt[3]{8.1}~=~2.00829$



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

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Saturday, October 12, 2024

22.5 - Solved Examples on Tangents And Normals

In the previous section, we saw Tangents and Normals. We saw some solved examples also. In this section, we will see a few more solved examples.

Solved example 22.18
Find the equation of the tangent to the curve
$\rm{y\,=\,\frac{x-7}{(x-2)(x-3)}}$ at the point where it cuts the x-axis.
Solution:
1. The derivative can be used to find the slope of tangent at any point. So we will first find the derivative.


2. We want the point at which the curve cuts the x-axis.
At that point, the y-coordinate will be zero. So in the equation of the curve, we substitute y by zero. We get:
$\rm{0\,=\,\frac{x-7}{(x-2)(x-3)}}$
⇒ x − 7 = 0
⇒ x = 7
• So the required point is: (7,0)

3. Next we want the slope at (7,0). We have:


4. So the equation of the tangent can be written as:
$\rm{y - y_0 ~=~m(x-x_0)}$
⇒ $\rm{y - 0 ~=~\frac{1}{20}(x-7)}$
⇒ 20y = x − 7
⇒ 20y − x + 7 = 0

Fig.22.18

• The graph is shown in fig.22.18 below:
    ♦ The curve is drawn in red color.
    ♦ The tangent is drawn in green color.


• The tangent is drawn at (7,0)
• The slope triangle has a height of 0.2 units and base of 4 units. So the slope of tangent is 0.2/4 = 1/20

Solved example 22.19
Find the equation of the tangent and normal to the curve
$\rm{x^{2/3} \,+\, y^{2/3}\,=\,2}$ at (1,1).
Solution:
1. The derivative can be used to find the slope of tangent at any point. So we will first find the derivative.


 

2. So we can write the slope of the tangent at (1,1):

$\rm{\left. \frac{dy}{dx} \right|_{(1,1)}~=~(-1) \left(\frac{1}{1} \right)^{1/3}~=~-1}$

3. Now we can write the equation of the tangent at (1,1).
y − y0 = m(x − x0)
⇒ y − 1 = (−1)(x − 1)
⇒ y − 1 = −x + 1
⇒ y + x − 2 = 0

4. Slope of the normal is equal to the negative reciprocal of that of the tangent. So slope of the normal is 1.
• Now we can write the equation of the tangent at (1,1).
y − y0 = m(x − x0)
⇒ y − 1 = (1)(x − 1)
⇒ y − 1 = x − 1
⇒ y − x = 0

• The graph is shown in fig.22.19 below:
    ♦ The curve is drawn in red color.
    ♦ The tangent is drawn in green color.

Fig.22.19


• The tangent is drawn at (1,1)
• The slope triangle has a height of −2 units and base of 2 units. So the slope of tangent is −2/2 = −1

Solved example 22.20
Find the equation of the tangent to the curve given by
$\rm{x \,=\, a \sin^3 t,~~y\,=\,b \cos^3 t}$ at  a point where t = π/2.
Solution:
1. The derivative can be used to find the slope of tangent at any point. So we will first find the derivative.


2. Next we want the slope at t = π/2. We have:
$\rm{\frac{dy}{dx}\,=\,\frac{-a \sin t}{b \cos t}}$
$\rm{~=\,\frac{-a \sin (\pi/2)}{b \cos (\pi/2)}}$
$\rm{~=\,\frac{-b \cos (\pi/2)}{a \sin (\pi/2)}}$
$\rm{~=\,\frac{-b (0)}{a (1)}}~=~0$

3. Next we want the (x,y) coordinates at t = π/2
$\rm{x \,=\, a \sin^3 (\pi/2)\,=\,a (1)^3 \,=\,a}$
$\rm{y\,=\,b \cos^3 (\pi/2)\,=\,b(0)^3 = 0}$

4. The slope of the tangent is zero. That means, the tangent is horizontal. So we can write the equation of the tangent just by using the y-coordinate obtained in (4).
• We get: y = 0

5. Fig.22.20 below shows the graph of the given function.
• It is assumed that, a = 3 and b = 4
• So the point (a,0) is (3,0)


Fig.22.20

• We see that:
If we draw the tangent at (3,0), it will be same as the x-axis.
• So the equation of the tangent is: y = 0


The link below gives a few more solved examples:

Exercise 22.3



In the next section, we will see Approximations.

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Thursday, September 26, 2024

22.4 - Tangents And Normals

In the previous section, we completed a discussion on the Derivative test which help us to find whether a given function is increasing or decreasing. In this section, we will see Tangents and Normals.

Some basic details can be written in 9 steps:
1. In analytic geometry classes, we saw that:
Equation of a straight line passing through (x0,y0) is given by
y – y0 = m(x – x0)
    ♦ Here m is the slope of that straight line.
2. So if we know the slope m, we can easily write the equation of the straight line passing through any given point (x0,y0).
3. Now consider a curve given by y = f(x).
• We have seen numerous curves of this type, in the previous sections.
• For example, f(x) = x3 – 3x2 + 3 is a curve.
• If we want to plot that curve on the xy-plane, we can write:
y = x3 – 3x2 + 3
4. Mark some random points on any given curve. Draw tangents through each of those points. We know that, no two tangents will be the same. This is because each tangent will have it’s own slope.
5. Suppose that, we want a tangent at a particular point (x0,y0).
• Based on (1) and (2) above, we can easily draw that tangent if we know the slope of the tangent at (x0,y0)
• But the slope of the tangent at (x0,y0) is f'(x0).
• So the equation of the required tangent is:
y – y0 = f'(x0)(x – x0)
6. Now we can write the equation of the normal also at (x0,y0).
• For that, we make use of the following fact:
Slope of the perpendicular line is the negative reciprocal of the slope of the original line.
• We can write:
Equation of the required normal is:
$\rm{y - y_0 \,=\,\frac{-1}{f'(x_0)} (x - x_0)}$
7. Suppose that, f'(x0) = 0
• Then it means that, the tangent at (x0,y0) is parallel to the x-axis.
• In such a situation, we can straight away write the equation of the tangent at (x0,y0) as: y = y0
8. Suppose that, f'(x0) tends to ∞.
• Then it means that, the tangent at (x0,y0) tends to be parallel to the y-axis.
• In such a situation, we can straight away write the equation of the tangent at (x0,y0) as: x = x0
9. The following information will be very useful while solving some types of problems. It can be written in 2 steps:
(i) If 𝜃 is the angle which a straight line makes with the +ve direction of the x-axis, then slope of that straight line will be equal to tan 𝜃. (We saw this in analytic geometry classes)
(ii) So we can write:
If the tangent at (x0,y0) makes an angle 𝜃 with the +ve direction of the x-axis, then f'(x0) = tan 𝜃.


Now we will see some solved examples:
Solved example 22.14
Find the slope of the tangent to the curve y = x3 − x at x = 2
Solution:
• Slope of the tangent at x = 2 is $\rm{\left. \frac{dy}{dx} \right |_{x = 2}}$
• It can be calculated as shown below:


• The graphs are shown in fig.22.14 below.
    ♦ The curve is drawn in red color.
    ♦ The tangent at x = 2 is drawn in green color.

Fig.22.14

• We see that, slope of the green line is 11.

Solved example 22.15
Find the point at which the tangent to the curve $\rm{y = \sqrt{4x – 3} ~-~1}$ has its slope $\rm{\frac{2}{3}}$.
Solution:
1. The derivative can be used to find the slope of tangent at any point. So we will first find the derivative.


2. The slope must be 2/3. So we can write:


3. Equation of the curve is: $\rm{y = \sqrt{4x – 3} ~-~1}$
• When x = 3, we get:
$\rm{y = \sqrt{4(3) – 3} ~-~1}~=~2$

4. Therefore, we can write:
Mark the point (3,2) on the given curve. Draw the tangent at that point. Slope of that tangent will be 2/3.

• The graph is shown in fig.22.15 below:
    ♦ The curve is drawn in red color.
    ♦ The tangent at (3,2) is drawn in green color.

Fig.22.15

• We see that, slope of the green line is 2/3 = 0.67

Note: In the above graph, we do not see much red curve below the x-axis. The reason can be written in 4 steps:
(i) When x = 3, we can write:
$\rm{y = \sqrt{4(3) – 3} ~-~1}~=~(\pm 3 -1)~=~2~\text{OR}~-4$
So (3, −4) is a possible point on the graph. But it is not plotted.
(ii) Another example:
When x = 7, we can write:
$\rm{y = \sqrt{4(7) – 3} ~-~1}~=~(\pm 5 -1)~=~4~\text{OR}~-6$
So (7, −6) is a possible point on the graph. But it is not plotted.
(iii) We do not plot such points. If we do, then it means that, for inputs like x = 3, 7 etc., there will be two outputs.
• If there are two outputs, we cannot call it a function.
(iv) So we restrict the output values (range).

Solved example 22.16
Find the equation of all lines having slope 2 and being tangent to the curve $\rm{y + \frac{2}{x - 3}~=~0}$.
Solution:
1. The derivative can be used to find the slope of tangent at any point. So we will first find the derivative.


2. The slope must be 2. So we can write:


3. Equation of the curve is: $\rm{y + \frac{2}{x - 3}~=~0}$
• When x = 4, we get:
$\rm{y = \frac{2}{3 - x}~=~\frac{2}{3 - 4}~=~-2}$
• When x = 2, we get:
$\rm{y = \frac{2}{3 - x}~=~\frac{2}{3 - 2}~=~2}$

4. Therefore, we can write:
There are two points (4,−2) and (2,2). Mark those two points on the given curve. Draw the tangent at each of those points. The two tangents will be parallel to each other with a slope of 2.

• The graph is shown in fig.22.16 below:
    ♦ The curve is drawn in red color.
    ♦ The tangents are drawn in green color.

Fig.22.16

• We see that, slope of the green lines is 2

5. Now we want the equation of the two tangents. We can use the general equation:
y – y0 = m(x – x0)
• So the equation of the tangent through (2,2) is:
y – 2 = 2(x – 2)
⇒ y − 2 = 2x − 4
⇒ y − 2x + 2 = 0
• Similarly, the equation of the tangent through (4,−2) is:
y – (−2) = 2(x – 4)
⇒ y + 2 = 2x − 8
⇒ y − 2x + 10 = 0

Solved example 22.17
Find points on the curve $\rm{\frac{x^2}{4}\,+\,\frac{y^2}{25}\,=\,1}$ at which tangents are (i) parallel to x-axis (ii) parallel to y-axis.
Solution:
• The derivative can be used to find the slope of tangent at any point. So we will first find the derivative.


Part (i): Tangents parallel to x-axis
1. Consider the points where tangents are parallel to the x-axis.
• Those tangents will have a slope of zero. So we can write:

2. When x = 0, we get:

3. So the points are: (0,5) and (0,−5)
The tangents at these points are parallel to the x-axis.

• The graph is shown in fig.22.16 below:
    ♦ The curve is drawn in red color.
    ♦ The tangents are drawn in green color.

Fig.22.17

Part (ii): Tangents parallel to y-axis
1. Consider the points where tangents are parallel to the y-axis.
• The normals at those points will have a slope of zero. Slope of normal is the negative reciprocal of that of tangent. So we can write:

2. When y = 0, we get:

3. So the points are: (2,0) and (−2,0)
• The normals at these points are parallel to the x-axis.
• Consequently, the tangents at these points are parallel to the y-axis. They are drawn in magenta color in fig.22.17 above.

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

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