Find-the-equation-of-tangent-and-normal-to-the-curve-y-given-by-y-x-3-3x-2-7-

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

F i n d t h e e q u a t i o n o f t a n g e n t a n d n o r m a l t o t h e c u r v e y

g i v e n b y y = x^{3} + 3 x^{2} + 7 .

Answered by MJS last updated on 20/Aug/18

f (x) = x^{3} + 3 x^{2} + 7

tangent in (\begin{matrix} p \\ f (p) \end{matrix})

y = k x + d \Rightarrow d = y - k x

k = f^{'} (p) = 3 p^{2} + 6 p

d = f (p) - f^{'} (p) p = - 2 p^{3} - 3 p^{2} + 7

t : y = (3 p^{2} + 6 p) x - 2 p^{3} - 3 p^{2} + 7

normal in (\begin{matrix} p \\ f (p) \end{matrix})

y = - \frac{1}{k} x + d \Rightarrow d = y + \frac{1}{k} x

k = same as above = f^{'} (p)

d = f (p) + \frac{1}{f^{'} (p)} p = p^{3} + 3 p^{2} + 7 + \frac{p}{3 p^{2} + 6 p} =

= p^{3} + 3 p^{2} + 7 + \frac{1}{3 p + 6} = \frac{(p^{3} + 3 p^{2} + 7) (3 p + 6) + 1}{3 p + 6} =

= \frac{3 p^{4} + 15 p^{3} + 18 p^{2} + 21 p + 43}{3 p + 6}

n : y = \frac{1}{3 (p + 2)} (- x + 3 p^{4} + 15 p^{3} + 18 p^{2} + 21 p + 43)

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