Question

Find the equations of the tangent and normal to the given curves at the indicated points:
(i) $y=x^4-6x^3+13x^2-10x+5$ at $(0,5)$.
(ii) $y=x^4-6x^3+13x^2-10x+5$ at $(1,3)$
(iii) $y=x^3$ at $(1,1)$
(iv) $y=x^2$ at $(0,0)$
(v) $x=\cos t,y=\sin t$ at $t=\frac{\pi }{4}$

Answer 1

(i)

We have
$y=x^4-6x^3+13x^2-10x+5$
$\Rightarrow \frac{dy}{dx}=4x^3-18x^2+26x-10$
$\therefore {\left(\frac{dy}{dx}\right)}_{(0,5)}=-10$
Thus, the slope of the tangent at $(0,5)$ is $-10$. Ergo equation of the tangent is given as,
$(y-5)=-10(x-0)\Rightarrow 10x+y-5=0$
The slope of the normal at $(0,5)$ is $\frac{-1}{\text{Slope of the tangent at}(0,5)}=\frac{1}{10}$.
Therefore, the equation of the normal at $(0,5)$ is given as,
$(y-5)=\frac{1}{10}(x-0)\Rightarrow 10y-50=x\Rightarrow x-10y+50=0$

(ii)

We have
$y=x^4-6x^3+13x^2-10x+5$
$\Rightarrow \frac{dy}{dx}=4x^3-18x^2+26x-10$
$\therefore {\left(\frac{dy}{dx}\right)}_{(1,3)}=4-18+26-10=2$
Thus, the slope of the tangent at $(1,3)$ is $2$. Ergo equation of the tangent is given as,
$(y-3)=2(x-1)\Rightarrow 2x-y+1=0$
The slope of the normal at $(1,3)$ is $\frac{-1}{\text{Slope of the tangent at}(1,3)}=-\frac{1}{2}$.
Therefore, the equation of the normal at $(1,3)$ is given as,
$(y-3)=-\frac{1}{2}(x-1)\Rightarrow x+2y-7=0$

(iii)

On differentiating with respect to $x$, we get:
$\frac{dy}{dx}=3x^2$
${\left(\frac{dy}{dx}\right)}_{(1,1)}=3(1{)}^2=3$
Thus, the slope of the tangent at $(1,1)$ is $3$ and the equation of the tangent is given as,
$y-1=3(x-1)\Rightarrow y=3x-2$
The slope of the normal at $(1,1)$ is $\frac{-1}{\text{Slope of the tangent at}(1,1)}=\frac{-1}{3}$.
Therefore, the equation of the normal at $(1,1)$ is given as,
$y-1=-\frac{1}{3}(x-1)\Rightarrow x+3y-4=0$

(iv)

We have
$y=x^2$
$\Rightarrow \frac{dy}{dx}=2x$
$\therefore {\left(\frac{dy}{dx}\right)}_{(0,0)}=0$
Thus, the slope of the tangent at $(0,0)$ is $0$. Ergo equation of the tangent is given as,
$(y-0)=0(x-0)\Rightarrow y=0$
The slope of the normal at $(0,0)$ is $\frac{-1}{\text{Slope of the tangent at}(0,0)}=-\infty$.
Therefore, the equation of the normal at $(0,0)$ is given as,
$(y-0)=-\infty (x-0)\Rightarrow x=0$