slope of the tangent line at $\left(x_0, y_0\right)$ is $\left.\frac{d y}{d x}\right|_{\left(x_0, y_0\right)}$ which equal $f^{\prime}\left(x_0\right)$ if $y=f(x)$

slope of the normal at ($x_0, y_0$) is = $\frac{-1}{\frac{dy}{dx}|_(x_0,y_0)}$

Recall: equ. of the line passing through $\left(x_0, y_0\right)$ and having slope m b given by

$y-y_0=m\left(x-x_0\right)$

$\therefore$ the equation of tangent at $P\left(x_0, y_1\right)$ is

$y-y_0=\left.\frac{d y}{d x}\right|_{\left(x_0, y_0\right)}\left(x-x_0\right)$.

The equ. of normal at $P\left(x_0, x\right)$ is

$y-y_0= \frac{-1}{\frac{dy}{dx}|_(x_0,y_0)} (x,x_0)$

if $.\frac{d y}{d x}|_{(x, y)} \neq 0$.

(1) Find the equation of the tangent line at a point $\left(x_0, y_0\right)$ on the circle $x^2+y^2=1$

Solution.

$x^2+y^2=1 \Rightarrow 2 x+2 y \frac{d y}{d x}=0$

$\Rightarrow \frac{d y}{d x}=-\frac{x}{y} \text { if } y \neq 0$

$\therefore$ If $y_0 \neq 0$ the slope of tangent at $\left(x_0, y_0\right)$ is $m=-\frac{x_0}{y_0}$.

$\therefore$ Equation of the tangent at $\left(x_0, y_0\right)$ is

$y-y_0=\frac{-x_0}{y_0}\left(x-x_0\right)$

$\Leftrightarrow y_0\left(y-y_0\right)+x_0\left(-x-x_0\right)=0$

$\Leftrightarrow \quad x_0 x+y_0 y=x_0^2+y_0^2=1$

$x_0 x+y_0 y = 1 ~$ if $~ y_0 \neq 0$

At the pts. $(1,0)$ and $(-1,0)$ the eqs. of the tangents are $x=1$ and $x=-1$, respectively.

conclusion: If the slope of the tangent line in infinite at be pt. $\left(x_0, y_0\right)$ then, the equation of the tangent is $x=x_0$

Find the point (s) at which the tangent to the curve

$y=\sqrt{4 x-3}-1$ has slope $2 / 3$.

Solution:

$\frac{d y}{d x}=\frac{1}{2 \sqrt{4 x-3}} \cdot 4=\frac{2}{\sqrt{4 x-3}}$

Slope of tangent at $(x, y)$ is $m=\frac{2}{\sqrt{4 x-3}}$

$\frac{2}{\sqrt{4 x-3}}=\frac{2}{3} \Rightarrow 4 x-3=9 \Rightarrow x=3$

when $x=3, y=\sqrt{4 \times 3-3}-1=2$

Hence, the required point is $(3,2)$.

Find points on the curve $\frac{x^2}{4}+\frac{y^2}{9}=1$ at which be tangents are

(i) parallel te be $x$-axis

(ii) parallel to the $y$-axis.

Solution. $\quad \frac{x^2}{4}+\frac{y^2}{9}=1$

$\begin{aligned}& \Rightarrow \quad \frac{2 x}{4}+\frac{2 y}{9} \frac{d y}{d x}=0 \\& \Rightarrow \quad \frac{d y}{d x}=-\frac{9}{4} \frac{x}{y}\end{aligned}$

For the tangents to be parallel to $x$-axis, the slope must be zero.

$\therefore$ We get $x=0$ for the tangent to be parallel to the x-axis.

Putting $x=0$ in the equ. of the curve,

we get $0+\frac{y^2}{9}=1 \Rightarrow \frac{y^2}{9}=1 \Rightarrow y= \pm 3$

So, (0,3) &(0,-3) are the points. Where the tangents the parallel to $x$-axis.

(ii) If the tangent is parallel to the $y$-axis, the. slope must be infinite.

$\therefore \quad y=0 \Rightarrow \frac{x^2}{4}=1 \Rightarrow x= \pm 2$

$\therefore ~$ (3,0) & (-,0) are the points where the tangent are parallel to the $y$ axis.

Find the equation of the tangent to the curve $y=\frac{x-7}{(x-2)(x-3)}$ at the point where it cuts the $x$-axis.

Solution:

Putting $y=0$, we get $x=7$;

So, the curve cuts tha $x$-axis at the point $(7,0)$.

$\frac{d y}{d x} =\frac{\frac{d}{d x}(x-7) \cdot(x-2)(x-3)-(x-7) \frac{d}{d x}[(x-2)(x-3)]}{(x-2)^2(x-3)^2}$

$=\frac{(x-2)(x-3)-(x-7)(2 x-5)}{(x-2)^2(x-3)^2}$

$\therefore$ The slope of the tangent at $(7,0)$ is

$\begin{aligned} m=\left.\frac{d y}{d x}\right|_{(7,0)} & \frac{(7-2)(7-3)-0}{(7-2)^2(7-3)^2} \\& =\frac{1}{5 \times 4}=\frac{1}{20}\end{aligned}$

$\therefore$ The equ. of the tangent at $(7,0)$ is $y-0=\frac{1}{20}(x-7)$

i.e. $~ 20 y=x-7$

Find the slope of the normal to the curve $x=a \cos ^3 \theta, y=a \sin ^3 \theta$ at the point where $\theta=\frac{\pi}{4}$

Solution :

$\frac{d x}{d \theta}=-3 a \cos ^2 \theta \sin \theta$

$\frac{d y}{d \theta}=3 a \sin ^2 \theta \cos \theta$

$\frac{d y}{d x}=\frac{d y / d \theta}{\frac{d x}{d \theta}}=\frac{3 a \sin ^2 \theta \cos \theta}{-3 a \cos ^2 \theta \sin \theta}=-\tan \theta$

The slope of tangent when $\theta=\frac{\pi}{4}$ is

$m=-\tan \frac{\pi}{4}=-1 \quad$

$\therefore$ slope of the normal =1

Find a point on the carve $y=(x-2)^2$ at which the tangent is parallel th the chord joining $(2,0)$ and $(4,4)$.

Solution:

Slope of the chord joining $(2,0)$ and $(4,4)$ is $m=\frac{4-0}{4-2}=2$.

$\therefore$ The slope of the tangent $=2$.

$y=(x-2)^2 \Rightarrow \frac{d y}{d x}=2(x-2)$

$2=2(x-2) \Rightarrow x-2=1 \Rightarrow x=3 \text {. }$

putting $x=3$, we get $y=(3-2)^2=1$

Hence, the point is $(3,1)$

If $\Delta x \rightarrow 0$, then $\frac{f(x+\Delta x)-f(x)}{\Delta x} \rightarrow f^{\prime}(x)$ So, if $\Delta x$ is small then the approximation $L(x+\Delta x)$ for $f(x+\Delta x)$ is not too bad.

The qu. of the tangent at $P(x, y)$ is give by

$Y-y=f^{\prime}(x)(X-x)$

$\therefore \quad Y=y+f^{\prime}(x)(X-x)$

At $X=x+\Delta x, \quad Y=yf^{\prime}(x) \Delta x$.

Hance, the linear approx is

$f(x+\Delta x) \approx f(x)+f^{\prime}(x) \Delta x$

Example 1: Approximate $\sqrt{36.6}$.

Sol. : $f(x)=\sqrt{x}, x=36, \Delta x=0.6$

Want $f(x+\Delta x)$.

$\begin{aligned} f(x+\Delta x) & \approx f(x)+f^{\prime}(x) \Delta x \\& =\sqrt{36}+\frac{1}{2 \sqrt{36}} \times 0.6 \\& =6+\frac{1}{20}=6.05\end{aligned}$

thus. $\sqrt{36.6} \approx 6.05$

Approx $(25)^{1 / 3}$.

Solution:

$f(x)=x^{1 / 3}, f^{\prime}(x)=\frac{1}{3} x^{-2 / 3}$

$x=27, x + \Delta x=25$

i.e.$~ \Delta x=-2$

$(25)^{1 / 3}=f(x+\Delta x)$

$\approx f(x)+f^{\prime}(x) \Delta x$

$=(2 x)^{1 / 3}+\frac{1}{3}(27)^{-2 / 3}(-2)$

$=3-\frac{2}{3 \times 9}=3-\frac{2}{27}=\frac{79}{27}\quad$

So, $(25)^{1 / 3} \approx \frac{79}{27}$

The radius of a sphere is measured as 9 cm with an error of 0.03 $\mathrm{~cm}$. Find the approximate even in volume.

Solution:

V $=\frac{4}{3} \pi r^3$

r $=9 \text { an }, \Delta r=0.03 \mathrm{~cm} .$

$\Delta V =\text { error in volume }$

$=V(r+\Delta r)-V(r) \approx V^{\prime}(r) \Delta r .$

$\therefore$ Approx. error in volume $=V^{\prime}(r) \Delta r$