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


Question and Answers Forum

All Questions Topic List
Others Questions
Previous in All Question Next in All Question
Previous in Others Next in Others
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

	$f (x) = x^{3} + 3 x^{2} + 7$ $tangent in (\begin{matrix} p \\ f (p) \end{matrix})$ $y = k x + d \Rightarrow d = y - k x$ $k = f^{'} (p) = 3 p^{2} + 6 p$ $d = f (p) - f^{'} (p) p = - 2 p^{3} - 3 p^{2} + 7$ $t : y = (3 p^{2} + 6 p) x - 2 p^{3} - 3 p^{2} + 7$ $normal in (\begin{matrix} p \\ f (p) \end{matrix})$ $y = - \frac{1}{k} x + d \Rightarrow d = y + \frac{1}{k} x$ $k = same as above = f^{'} (p)$ $d = f (p) + \frac{1}{f^{'} (p)} p = p^{3} + 3 p^{2} + 7 + \frac{p}{3 p^{2} + 6 p} =$ $= p^{3} + 3 p^{2} + 7 + \frac{1}{3 p + 6} = \frac{(p^{3} + 3 p^{2} + 7) (3 p + 6) + 1}{3 p + 6} =$ $= \frac{3 p^{4} + 15 p^{3} + 18 p^{2} + 21 p + 43}{3 p + 6}$ $n : y = \frac{1}{3 (p + 2)} (- x + 3 p^{4} + 15 p^{3} + 18 p^{2} + 21 p + 43)$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum