  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,$$(y-0)=-\infty(x-0)\Rightarrow x=0$$MathematicsRS AgarwalStandard XII

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