Application of Derivatives

y = 5x² + 2x – 25

P(2, -1)

T_(p) : T = 0

->y – 1 = 10x(2) + 2(x + 2) – 50

$\Rightarrow y=22x-45$ is also tangent to y = x³ – x² + x at point (a, b)

For y = x³ -x² + x

$\frac{dy}{dx}=3x^2-2x+1>0$

y = 22x  $-\frac{1939}{27}$ which is not tangent to the curve.

$3a^2-2a+1=22$  (slope of tangent)

$\Rightarrow 3a^2-2a-21=0\to a=\frac{2\pm \sqrt[]{4+252}}{6}=\frac{2\pm 16}{6}=3,\frac{-7}{3}$

b = 27 – 9 + 3 = 21

tangent : y – 21 = 22(x – 3)

->y = 22x – 45

a = 3, b = 21

2a + 9b = 6 + 189 = 195

Also,   $a^3-a^2+a=b$

For a = $\frac{-7}{3}$  

b =   $\frac{-343}{27}-\frac{49}{9}-\frac{7}{3}$

=   $\frac{-343-147-63}{27}=\frac{-553}{27}$

Common tangents

$T_1:y=mx\pm \sqrt[]{4m^2+9}$      

$T_2:y=mx\pm \sqrt[]{42m^2-13}$    

So, 4m² + 9 = 42 m² – 13 Þ m =   $\pm 2$

  $\Rightarrow c=\pm 5$              

So tangent  $y=\pm 2x\pm 5$

y = 2x + 5 does not pass through 4^th quadrant compare this tangent with T = 0 to get pt. of intersection

$\frac{x{x}_2}{42}-\frac{y{y}_2}{143}=1\Rightarrow x_2=\frac{-84}{5}$

Find a, α

$\left({\alpha }^2+{\alpha }^2-3\right)+{\left({\alpha }^2-{\alpha }^2-1\right)}^5=0$                

$\alpha =\sqrt[]{2}$      

zx differentiate the curve

$2x+2y.y^1+5{\left(x^2-y^2-1\right)}^4\left(2x-2y\right)=0$ ……. (i)

differentiate equation (i)

$\left(\alpha ,\alpha \right)\Rightarrow y\text{'}\text{'}=\frac{-23}{4\sqrt[]{2}}$          

$V=\frac{1}{3}\pi r^{2}h$

$V=\frac{\pi }{8}{h}^{3}ta{n}^2\alpha$

$\frac{dv}{dt}=\pi h^{2}ta{n}^2\alpha .\frac{dh}{dt}$

CSA = $\pi r\mathcal{l}$ $$

$\frac{d\left(CSA\right)}{dt}=\frac{15}{16}\pi 2h.\frac{dh}{dt}$

$\frac{15}{8}*\frac{2}{3}=5m^2/hrs$

y⁵ – 9xy + 2x = 0

differentiate 5y⁴ – 9x $\frac{dy}{dx}$ 9y + 2 = 0

$\frac{dy}{dx}=\frac{9y-2}{5y^4-9x}$

For horizontal tangent $\frac{dy}{dx}=0\Rightarrow y=\frac{2}{9}$ which does not satisfy the equation so no horizontal

For vertical tangent $5y^4-9x=0$

m = 0, N = 2

$4x^3-3x{y}^2+6x^2-5xy-8y^2+9x+14=0$ differentiating both sides we get

$12x^2-3y^2-6xyy\text{'}+12x-5y-5xy=16yy\text{'}+9=0$

At the point (2, 3)

48 – 27 + 36y' – 24 – 15 + 10y' – 48y' + 9 = 0

$\therefore Area=\frac{1}{2}*Base*Height$

$A=\frac{1}{2}*\left(\frac{-4}{3}+\frac{31}{2}\right)\left(3\right)=\frac{1}{2}\left(\frac{85}{0}\right).3=\frac{85}{4}=8A=170$

$f\left(x\right)+{\displaystyle {\int }_0^x\left(x-t\right)f\text{'}\left(t\right)dt=\left(e^{2x}+e^{-2x}\right)cos2x+\frac{2x}{a}.......(i)}$

Here f (0) = 2 ………. (ii)

On differentiating equation (i) w.r.t. x we get :

$f\text{'}\left(x\right)+f{\displaystyle {\int }_0^{x}f\text{'}\left(t\right)dt+xf\text{'}\left(x\right)-xf\text{'}\left(x\right)}$

$4=\frac{2}{a}\Rightarrow a=\frac{1}{2}$

$\therefore {\left(2a+1\right)}^5.a^2=2^5.\frac{1}{2^2}=2^3=8$

y = 5x² + 2x – 25

P(2, -1)

T_(p) : T = 0

Þ y – 1 = 10x(2) + 2(x + 2) – 50

$\Rightarrow y=22x-45$ is also tangent to y = x³ – x² + x at point (a, b)

For y = x³ -x² + x

$\frac{dy}{dx}=3x^2-2x+1>0$               

y = 22x $-\frac{1939}{27}$  which is not tangent to the curve.

$3a^2-2a+1=22$ (slope of tangent)

$\Rightarrow 3a^2-2a-21=0\to a=\frac{2\pm \sqrt[]{4+252}}{6}=\frac{2\pm 16}{6}=3,\frac{-7}{3}$               

b = 27 – 9 + 3 = 21

tangent : y – 21 = 22(x – 3)

 ⇒ y = 22x – 45

a = 3, b = 21

2a + 9b = 6 + 189 = 195

Also, $a^3-a^2+a=b$  

For a = $\frac{-7}{3}$  

b = $\frac{-343}{27}-\frac{49}{9}-\frac{7}{3}$  

= $\frac{-343-147-63}{27}=\frac{-553}{27}$  

$f\text{'}\left(a\right)=f\text{'}\left(b\right)=f\text{'}\left(c\right)=2$

f'' (x) is zero for atleast $x_1\in \left(a,b\right)\&x_2\in \left(b,c\right)$ ${}_{}{}_{}$

$\left(x+1\right)\frac{dy}{dx}-y=e^{3x}{\left(x+1\right)}^2,$

$\left(x+1\right)dy-ydx=e^{3x}{\left(x+1\right)}^{2}dx$

$\frac{\left(x+1\right)dy-ydx}{{\left(x+1\right)}^2}=e^{3x}dx$

$\Rightarrow d\left(\frac{y}{x+1}\right)=e^{3x}dx\Rightarrow \frac{y}{x+1}=\frac{e^{3x}}{3}+c$

$=\frac{e^{3x}}{3}\left(3x+4\right)$