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

Question:

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

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

(ii) $y=x^{4}-6 x^{3}+13 x^{2}-10 x+5$ 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 $t=\frac{\pi}{4}$

Solution:

(i) The equation of the curve is $y=x^{4}-6 x^{3}+13 x^{2}-10 x+5$.

On differentiating with respect to x, we get:

$\frac{d y}{d x}=4 x^{3}-18 x^{2}+26 x-10$

$\left.\frac{d y}{d x}\right]_{(0,5)}=-10$

Thus, the slope of the tangent at (0, 5) is −10. The equation of the tangent is given as:

$y-5=-10(x-0)$

$\Rightarrow y-5=-10 x$

$\Rightarrow 10 x+y=5$

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:

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

$\Rightarrow 10 y-50=x$

$\Rightarrow x-10 y+50=0$

(ii) The equation of the curve is $y=x^{4}-6 x^{3}+13 x^{2}-10 x+5$.

On differentiating with respect to x, we get:

$\frac{d y}{d x}=4 x^{3}-18 x^{2}+26 x-10$

$\left.\frac{d y}{d x}\right]_{(1,3)}=4-18+26-10=2$

Thus, the slope of the tangent at (1, 3) is 2. The equation of the tangent is given as:

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

$\Rightarrow y-3=2 x-2$

$\Rightarrow y=2 x+1$

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:

$y-3=-\frac{1}{2}(x-1)$

$\Rightarrow 2 y-6=-x+1$

$\Rightarrow x+2 y-7=0$

(iii) The equation of the curve is $y=x^{3}$.

On differentiating with respect to x, we get:

$\frac{d y}{d x}=3 x^{2}$

$\left.\frac{d y}{d x}\right]_{(1,1)}=3(1)^{2}=3$

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

$y-1=3(x-1)$

$\Rightarrow y=3 x-2$

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:

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

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

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

(iv) The equation of the curve is $y=x^{2}$.

On differentiating with respect to x, we get:

$\frac{d y}{d x}=2 x$

$\left.\frac{d y}{d x}\right]_{(0,0)}=0$

Thus, the slope of the tangent at (0, 0) is 0 and the equation of the tangent is given as:

y − 0 = 0 (x − 0)

$\Rightarrow y=0$

The slope of the normal at $(0,0)$ is $\frac{-1}{\text { Slope of the tangent at }(0,0)}=-\frac{1}{0}$, which is not defined.

Therefore, the equation of the normal at (x0, y0) = (0, 0) is given by

(v) The equation of the curve is x = cos ty = sin t.

$x=\cos t$ and $y=\sin t$

$\therefore \frac{d x}{d t}=-\sin t, \frac{d y}{d t}=\cos t$

$\therefore \frac{d y}{d x}=\frac{\left(\frac{d y}{d t}\right)}{\left(\frac{d x}{d t}\right)}=\frac{\cos t}{-\sin t}=-\cot t$

$\left.\frac{d y}{d x}\right]_{t=\frac{\pi}{4}}=-\cot t=-1$

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

When $t=\frac{\pi}{4}, x=\frac{1}{\sqrt{2}}$ and $y=\frac{1}{\sqrt{2}}$.

Thus, the equation of the tangent to the given curve at $t=\frac{\pi}{4}$ i.e., at $\left[\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\right]$ is

$y-\frac{1}{\sqrt{2}}=-1\left(x-\frac{1}{\sqrt{2}}\right)$

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

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

The slope of the normal at $t=\frac{\pi}{4}$ is $\frac{-1}{\text { Slope of the tangent at } t=\frac{\pi}{4}}=1$.

Therefore, the equation of the normal to the given curve at $t=\frac{\pi}{4}$ i.e., at $\left[\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\right]$ is

$y-\frac{1}{\sqrt{2}}=1\left(x-\frac{1}{\sqrt{2}}\right)$

$\Rightarrow x=y$