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Question Number 42207 by Rio Michael last updated on 20/Aug/18

Find the equation of tangent and normal to  the curve y  given by   y = x^3  + 3x^2  + 7 .

$${Find}\:{the}\:{equation}\:{of}\:{tangent}\:{and}\:{normal}\:{to}\:\:{the}\:{curve}\:{y} \\ $$$${given}\:{by}\:\:\:{y}\:=\:{x}^{\mathrm{3}} \:+\:\mathrm{3}{x}^{\mathrm{2}} \:+\:\mathrm{7}\:. \\ $$

Answered by MJS last updated on 20/Aug/18

f(x)=x^3 +3x^2 +7  tangent in  ((p),((f(p))) )  y=kx+d ⇒ d=y−kx  k=f′(p)=3p^2 +6p  d=f(p)−f′(p)p=−2p^3 −3p^2 +7    t: y=(3p^2 +6p)x−2p^3 −3p^2 +7    normal in  ((p),((f(p))) )  y=−(1/k)x+d ⇒ d=y+(1/k)x  k=same as above=f′(p)  d=f(p)+(1/(f′(p)))p=p^3 +3p^2 +7+(p/(3p^2 +6p))=  =p^3 +3p^2 +7+(1/(3p+6))=(((p^3 +3p^2 +7)(3p+6)+1)/(3p+6))=  =((3p^4 +15p^3 +18p^2 +21p+43)/(3p+6))    n: y=(1/(3(p+2)))(−x+3p^4 +15p^3 +18p^2 +21p+43)

$${f}\left({x}\right)={x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} +\mathrm{7} \\ $$$$\mathrm{tangent}\:\mathrm{in}\:\begin{pmatrix}{{p}}\\{{f}\left({p}\right)}\end{pmatrix} \\ $$$${y}={kx}+{d}\:\Rightarrow\:{d}={y}−{kx} \\ $$$${k}={f}'\left({p}\right)=\mathrm{3}{p}^{\mathrm{2}} +\mathrm{6}{p} \\ $$$${d}={f}\left({p}\right)−{f}'\left({p}\right){p}=−\mathrm{2}{p}^{\mathrm{3}} −\mathrm{3}{p}^{\mathrm{2}} +\mathrm{7} \\ $$$$ \\ $$$${t}:\:{y}=\left(\mathrm{3}{p}^{\mathrm{2}} +\mathrm{6}{p}\right){x}−\mathrm{2}{p}^{\mathrm{3}} −\mathrm{3}{p}^{\mathrm{2}} +\mathrm{7} \\ $$$$ \\ $$$$\mathrm{normal}\:\mathrm{in}\:\begin{pmatrix}{{p}}\\{{f}\left({p}\right)}\end{pmatrix} \\ $$$${y}=−\frac{\mathrm{1}}{{k}}{x}+{d}\:\Rightarrow\:{d}={y}+\frac{\mathrm{1}}{{k}}{x} \\ $$$${k}=\mathrm{same}\:\mathrm{as}\:\mathrm{above}={f}'\left({p}\right) \\ $$$${d}={f}\left({p}\right)+\frac{\mathrm{1}}{{f}'\left({p}\right)}{p}={p}^{\mathrm{3}} +\mathrm{3}{p}^{\mathrm{2}} +\mathrm{7}+\frac{{p}}{\mathrm{3}{p}^{\mathrm{2}} +\mathrm{6}{p}}= \\ $$$$={p}^{\mathrm{3}} +\mathrm{3}{p}^{\mathrm{2}} +\mathrm{7}+\frac{\mathrm{1}}{\mathrm{3}{p}+\mathrm{6}}=\frac{\left({p}^{\mathrm{3}} +\mathrm{3}{p}^{\mathrm{2}} +\mathrm{7}\right)\left(\mathrm{3}{p}+\mathrm{6}\right)+\mathrm{1}}{\mathrm{3}{p}+\mathrm{6}}= \\ $$$$=\frac{\mathrm{3}{p}^{\mathrm{4}} +\mathrm{15}{p}^{\mathrm{3}} +\mathrm{18}{p}^{\mathrm{2}} +\mathrm{21}{p}+\mathrm{43}}{\mathrm{3}{p}+\mathrm{6}} \\ $$$$ \\ $$$${n}:\:{y}=\frac{\mathrm{1}}{\mathrm{3}\left({p}+\mathrm{2}\right)}\left(−{x}+\mathrm{3}{p}^{\mathrm{4}} +\mathrm{15}{p}^{\mathrm{3}} +\mathrm{18}{p}^{\mathrm{2}} +\mathrm{21}{p}+\mathrm{43}\right) \\ $$

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