Question

Find the equations of the tangent and normal to the given curves at the given points

(i) $y=x^4-6x^3+13x^2-10x+5at(0,5)$

(ii) $y=x^4-6x^3+13x^2-10x+5at(1,3)$

(iii) $y=x^{3}at(1,1)$

(iv) $y=x^{2}at(0,0)$

(v) $x=cost,y=sint=\frac{\pi }{4}$

Answer 1

(i) The equation of the given curve is,

$y=x^4-6x^3+13x^2-10x+5$

On differentiating w.r.t. x, we get $\frac{dy}{dx}=4x^3-18x^2+26x-10$

$\therefore$ Slope of tangent at (0,5) is ${\left(\frac{dy}{dx}\right)}_{(0,5)}$ =0-0+0-10=-10

Thus, the slope of the tangent at (0,5) is -10. Now, the equation of the tangent is $y-5=-10(x-0)\Rightarrow 10x+y-5=0$

Again, the slope of normal at (0,5) is

$\frac{-1}{Slopeoftangentat(0,5)}=\frac{-1}{-10}=\frac{1}{10}$

Therefore, the equation of the normal at (0,5) is

$y-5=\frac{1}{10}(x-0)\Rightarrow x-10y+50=0$

(ii) The equation of the given curve is,

$y=x^4-6x^3+13x^2-10x+5$

On differentiating w.r.t. x, we get $\frac{dy}{dx}=4x^3-18x^2+26x-10$

$\therefore$ Slope of tangent at (1,3) is ${\left(\frac{dy}{dx}\right)}_{(1,3)}$ =14-18+26-10=2

Thus, the slope of the tangent at (1,3) is 2. Now, the equation of the tangent is

$y-3=2(x-1)\Rightarrow y-3=2x-2\Rightarrow y=2x+1$

Again, the slope of normal at (1,3) is

$\frac{-1}{Slopeoftangentat(1,3)}=\frac{-1}{2}$

Hence, the equation of the normal at (1,3) is

$y-3=\frac{-1}{2}(x-1)\Rightarrow 2y-6=-x+1\Rightarrow x+2y-7=0$

(iii) The equation of the given curve is $y=x^3$

On differentiating w.r.t. x, we get $\frac{dy}{dx}=3x^2$

$\therefore$ Slope of tangent at (1,1) is ${\left(\frac{dy}{dx}\right)}_{(1,1)}=3(1{)}^2=3$

Thus, the slope of the tangent at (1,1) is 3 and the equation of the tangent is

$y-1=3(x-1)\Rightarrow y-1=3x-3\Rightarrow y=3x-2$

Again, the slope of normal at (1,1)

=$\frac{-1}{Slopeoftangentat(1,1)}=\frac{-1}{3}$

Hence, the equation of the normal at (1,1) is

$y-1=\frac{-1}{3}(x-1)$

$\Rightarrow 3y-3=-x+1\Rightarrow x+3y-4=0$

(iv) The equation of the given curve is $y=x^2$

On differentiating w.r.t. x, we get $\frac{dy}{dx}=2x$

$\therefore$ Slope of tangent at (0,0) is ${\left(\frac{dy}{dx}\right)}_{(0,0)}=2\times 0=0$

Hence, the slope of the tangent at (0,0) having slope 0 is

$y-0=0(x-0)\Rightarrow y=0$

the slope of normal at (0,0) is $\frac{-1}{Slopeoftangentat(0,0)}=\frac{1}{0}$

Which is not defined.

Therefore, the equation of the normal at (x_0,y_0)=(0,0) is given by

$x=x_0=0$

(v) The equation of the given curve is $x=costy=sint$ ...(i)

The Point on the curve corresponding to $t=\frac{\pi }{4}$ is

$x=cos\frac{\pi }{4},y=sin\frac{\pi }{4}$ $(Onputtingt=\frac{\pi }{4})$

$\Rightarrow x=\frac{1}{\sqrt{2}},y=\frac{1}{\sqrt{2}}$

On differentiating w.r.t. x, we get

$\frac{dx}{dt}=-sint,\frac{dy}{dt}=cost\therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{cost}{-sint}=-$cot t

${\left(\frac{dy}{dx}\right)}_{t=\frac{\pi }{4}}=-cot\frac{\pi }{4}=-1$

$\therefore$ Slope of tangent at $t=\frac{\pi }{4}$ is -1.

Hence, the equation of the tangent to the given curve at $t=\frac{\pi }{4}$ i.e.,

at $(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})isy-\frac{1}{\sqrt{2}}=-(x-\frac{1}{\sqrt{2}})$

$\Rightarrow x+y-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=0\Rightarrow x+y-\frac{2}{\sqrt{2}}=0\Rightarrow x+y-\sqrt{2}=0$

Again, the slope of normal at $t=\frac{\pi }{4}$ is

$\frac{-1}{Slopeoftangentat=\frac{\pi }{4}}=\frac{-1}{-1}=1$

Hence, the equation of the normal to the given curve at $t=\frac{\pi }{4}$

i.e.,at $(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})isy-\frac{1}{\sqrt{2}}=1(x-\frac{1}{\sqrt{2}})\Rightarrow x-y=0$.