mathematics-class-12-unit-06-chapter-11-derivatives-l-11-12-qv6f-p9ctke

### <Success style="font-size:25px"> Derivatives L-11 </Success>
### <primary style="font-size:24px"> Equations of tangents and normals </primary>
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- 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 = $\frac{-1}{\frac{dy}{dx}|_(x_0,y_0)}$

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<footer style = "font-size: 18px"> $\rightarrow$ Derivatives lecture-11 $\rightarrow$ <span style = "color:blue"> Equations of tangents and normals </span> $\rightarrow$ Equations of tangents and normals $\rightarrow$ Problem-1 </footer>

### <Success style="font-size:25px"> Derivatives L-11 </Success>
### <primary style="font-size:24px"> Equations of tangents and normals </primary>
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- 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 $.\frac{d y}{d x}|_{(x, y)} \neq 0$.

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<footer style = "font-size: 18px">Derivatives lecture-11 $\rightarrow$ Equations of tangents and normals $\rightarrow$ <span style = "color:blue"> Equations of tangents and normals </span> $\rightarrow$ Problem-1 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Derivatives L-11 </Success>
### <primary style="font-size:24px"> Problem-1 </primary>
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- (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)$
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<footer style = "font-size: 18px">Equations of tangents and normals $\rightarrow$ Equations of tangents and normals $\rightarrow$ <span style = "color:blue"> Problem-1 </span> $\rightarrow$ Solution $\rightarrow$ Problem-2 </footer>

### <Success style="font-size:25px"> Derivatives L-11 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $\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 $x=x_0$

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<footer style = "font-size: 18px">Equations of tangents and normals $\rightarrow$ Problem-1 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-2 $\rightarrow$ Problem-3 </footer>

### <Success style="font-size:25px"> Derivatives L-11 </Success>
### <primary style="font-size:24px"> Problem-2 </primary>
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- 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 $(3,2)$.

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<footer style = "font-size: 18px">Problem-1 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-2 </span> $\rightarrow$ Problem-3 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Derivatives L-11 </Success>
### <primary style="font-size:24px"> Problem-3 </primary>
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- 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.
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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-2 $\rightarrow$ <span style = "color:blue"> Problem-3 </span> $\rightarrow$ Solution $\rightarrow$ Problem-4 </footer>

### <Success style="font-size:25px"> Derivatives L-11 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $\therefore$ 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.

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<footer style = "font-size: 18px">Problem-2 $\rightarrow$ Problem-3 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-4 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Derivatives L-11 </Success>
### <primary style="font-size:24px"> Problem-4 </primary>
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- 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} $

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<footer style = "font-size: 18px">Problem-3 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-4 </span> $\rightarrow$ Solution $\rightarrow$ Problem-5 </footer>

### <Success style="font-size:25px"> Derivatives L-11 </Success>
### <primary style="font-size:24px"> Problem-5 </primary>
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- 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

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<footer style = "font-size: 18px">Problem-4 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-5 </span> $\rightarrow$ Problem-6 $\rightarrow$ Application of the tangent line to approximation </footer>

### <Success style="font-size:25px"> Derivatives L-11 </Success>
### <primary style="font-size:24px"> Problem-6 </primary>
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- 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 $(3,1)$

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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-5 $\rightarrow$ <span style = "color:blue"> Problem-6 </span> $\rightarrow$ Application of the tangent line to approximation $\rightarrow$ Application of the tangent line to approximation </footer>

### <Success style="font-size:25px"> Derivatives L-11 </Success>
### <primary style="font-size:24px"> Application of the tangent line to approximation </primary>
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- Suppose it is easy to compute $f(x)$ but not so easy to compute $f(x+\Delta x)$. we would like to approximate $f(x+\Delta x)$ by some value where is easy to compute.

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<footer style = "font-size: 18px">Problem-5 $\rightarrow$ Problem-6 $\rightarrow$ <span style = "color:blue"> Application of the tangent line to approximation </span> $\rightarrow$ Application of the tangent line to approximation $\rightarrow$ Problem-7 </footer>

### <Success style="font-size:25px"> Derivatives L-11 </Success>
### <primary style="font-size:24px"> Application of the tangent line to approximation </primary>
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- 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$

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<footer style = "font-size: 18px">Problem-6 $\rightarrow$ Application of the tangent line to approximation $\rightarrow$ <span style = "color:blue"> Application of the tangent line to approximation </span> $\rightarrow$ Problem-7 $\rightarrow$ Problem-8 </footer>

### <Success style="font-size:25px"> Derivatives L-11 </Success>
### <primary style="font-size:24px"> Problem-7 </primary>
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- 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. $\sqrt{36.6} \approx 6.05$

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<footer style = "font-size: 18px">Application of the tangent line to approximation $\rightarrow$ Application of the tangent line to approximation $\rightarrow$ <span style = "color:blue"> Problem-7 </span> $\rightarrow$ Problem-8 $\rightarrow$ Problem-9 </footer>

### <Success style="font-size:25px"> Derivatives L-11 </Success>
### <primary style="font-size:24px"> Problem-8 </primary>
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- 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, $(25)^{1 / 3} \approx \frac{79}{27}$

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<footer style = "font-size: 18px">Application of the tangent line to approximation $\rightarrow$ Problem-7 $\rightarrow$ <span style = "color:blue"> Problem-8 </span> $\rightarrow$ Problem-9 $\rightarrow$ Thank you </footer>

### <Success style="font-size:25px"> Derivatives L-11 </Success>
### <primary style="font-size:24px"> Problem-9 </primary>
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- 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 $=V^{\prime}(r) \Delta r$

- $ =4 \pi \gamma^2 \Delta r $
- $ =4 \pi(7)^2 \times 0.03 \mathrm{~cm}^3 .$

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<footer style = "font-size: 18px">Problem-7 $\rightarrow$ Problem-8 $\rightarrow$ <span style = "color:blue"> Problem-9 </span> $\rightarrow$ Thank you $\rightarrow$  </footer>