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

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

(ii) $y = {x^4} - d + 13{x^2} - 10x + 5\;$at $(1,\;3)$

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

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

(v) $x = \cos t,y = \sin t$ at $t = \cfrac{\pi }{4}.$

(i) We have,
$y = {x^4} - 6{x^3} + 13{x^2} - 10x + 5$

…(1)

Differentiating (1) w.r.t. $x$, we get $\cfrac{{dy}}{{dx}} = 4{x^3} - 18{x^2} + 26x - 10$

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

So, equation of the tangent to (1) at $(0,5)$ is
$y - 5 = - 10(x - 0)$ , or $10x + y - 5 = 0$

Again, the slope of normal at $(0,5)$
$= \cfrac{{ - 1}}{{{\rm{Slope}}\,\,{\rm{of}}\,\,{\rm{tangent}}}} = \cfrac{{ - 1}}{{ - 10}} = \cfrac{1}{{10}}.$

So, equation of the normal to (1) at $(0,5)$ is
$y - 5 = \cfrac{1}{{10}}(x - 0)$ , or $x - 10y + 50 = 0$

(ii) We have,
$y = {x^4} - 6{x^3} + 13{x^2} - 10x + 5$ …(1)

Differentiating (1) w.r.t. $x$, we get$\cfrac{{dy}}{{dx}} = 4{x^3} - 18{x^2} + 26x - 10$

Therefore the slope of tangent at $(1,\;3)$ is ${\left( {\cfrac{{dy}}{{dx}}} \right)_{(1,3)}} = 4 - 18 + 26 - 10 = 2$

So, equation of the tangent to (1) at $(1,\;3)$ is
$y - 3 = 2(x - 1)$ or $2x - y + 1 = 0$

Again, the slope of normal at $(1,\;3) = \cfrac{{ - 1}}{{{\rm{Slope}}\,\,{\rm{of}}\,\,{\rm{tangent}}}} = \cfrac{{ - 1}}{2}$

Hence, the equation of the normal to (1) at $(1,\;3)$ is
$y - 3 = - \cfrac{1}{2}(x - 1)$ , or $x + 2y - 7 = 0$
…(1)

(iii) We have,
$y = {x^3}$ …(1)

Differentiating (1) w.r.t. $x$, we get $\cfrac{{dy}}{{dx}} = 3{x^2}$

So the slope of the tangent to (1) at $(1,\;1)$ is ${\left( {\cfrac{{dy}}{{dx}}} \right)_{(1,1)}} = 3{(1)^2} = 3$

Therefore the equation of the tangent to (1) at $(1,\;1)$ is
$y - 1 = 3(x - 1)$ or $3x - y - 2 = 0$

Similarly, the slope of normal at $(1,\;1) = \cfrac{{ - 1}}{{{\rm{Slope}}\,\,{\rm{of}}\,\,{\rm{tangent}}}} = \cfrac{{ - 1}}{3}$

Therefore the equation of the normal to$(1)$ at $(1,1)$is
$y - 1 = - \cfrac{1}{3}(x - 1)$ or $x + 3y - 4 = 0$

(iv) The given curve is $y = {x^2}$ …(1)
Differentiating (1) w.r.t. $x$, we get $\cfrac{{dy}}{{dx}} = 2x$

The slope of the tangent to (1) at $\left( {0,0} \right) = {\left( {\cfrac{{dy}}{{dx}}} \right)_{\left( {0,0} \right)}} = 2 \times 0 = 0$

So , the equation of the tangent to (1) at $(0,0)$ is
$y - 0 = 0(x - 0)$ or $y = 0$

The equation of the normal line to (1) at $(0,0)$ is
$(y - 0) = - \cfrac{1}{{{{\left( {\cfrac{{dy}}{{dx}}} \right)}_{(0,0)}}}}(x - 0)$

$y{\left( {\cfrac{{dy}}{{dx}}} \right)_{\left( {0,0} \right)}} = - x \Rightarrow y\left( 0 \right) = - x \Rightarrow x = 0$

Alternately, tangent at $(0,0)$ is parallel to $x$-axis, Therefore, normal to (1) at $(0,0)$is parallel to $y$-axis and its equation is $x = 0.$
( Line through $(0,0)$ and parallel to $y$-axis is $x = 0$)

(v) We have,
$x = \cos t,\,y = \sin t$ …(1)
$\Rightarrow \cfrac{{dx}}{{dt}} = - \sin t,\cfrac{{dy}}{{dt}} = \cos t$

The point on the curve at $t = \cfrac{\pi }{4}$ is
$\left( {\cos \cfrac{\pi }{4},\;\sin \cfrac{\pi }{4}} \right)$ or $\left( {\cfrac{1}{{\sqrt 2 }},\;\cfrac{1}{{\sqrt 2 }}} \right)$

Also, the slope of the tangent at $t$
$= \cfrac{{\left( {\cfrac{{dy}}{{dt}}} \right)}}{{\left( {\cfrac{{dx}}{{dt}}} \right)}} = \cfrac{{\cos t}}{{ - \sin t}} = - \cot t$

Therefore the slope of the tangent to (1) at $t = \cfrac{\pi }{4}$is $- \cot (\pi /4) = - 1.$

So, the equation of the tangent to (1) at $t = \cfrac{\pi }{4}$ is
$y - \cfrac{1}{{\sqrt 2 }} = - 1\left( {x - \cfrac{1}{{\sqrt 2 }}} \right)$ or $x + y - \cfrac{2}{{\sqrt 2 }} = 0$
or $x + y - \sqrt 2 = 0$

Also, the slope of the normal to (1) at $t = \cfrac{\pi }{4}$ is
$\cfrac{{ - 1}}{{{\rm{Slope}}\,\,{\rm{of}}\,\,{\rm{tangent}}}} = \cfrac{{ - 1}}{{ - 1}} = 1$

Therefore the equation of the normal to (1) at $t = \cfrac{\pi }{4}$ is
$y - \cfrac{1}{{\sqrt 2 }} = 1\left( {x - \cfrac{1}{{\sqrt 2 }}} \right)$ or $x - y = 0$