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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 $$t=\dfrac{\pi}4$$


Solution

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

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

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

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

Mathematics
RS Agarwal
Standard XII

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