My Math Forum Curve with gradient function

 Calculus Calculus Math Forum

 January 8th, 2014, 11:22 PM #1 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 Curve with gradient function A curve with gradient function 3x^2-12x+9 passes through the point M(2,-4) Find the gradient of the curve at M The equation of the normal curve at M The equation of the curve
 January 8th, 2014, 11:56 PM #2 Senior Member   Joined: Dec 2012 Posts: 372 Thanks: 2 Re: Curve with gradient function $y'(x)= 3x^2 - 12x +9 \ \Rightarrow \ y#39;(2) = -3$. This is the gradient at M. To find the equation of the normal to the curve at M, we first note that the normal vector must be perpendicular to the tangent vector. Since the tangent has slope of -3, the normal must have a slope of $\frac{1}{3}.$ The normal has equation $\dfrac{y + 4}{x - 2}= \dfrac{1}{3}$. $y(x)= \int (3x^2 - 12x +9)dx = x^3 - 6x^2 + 9x + k$ $y(2)= -4 \ \Rightarrow \ 8 - 24 + 18 + k = - 4 \ \Rightarrow \ k = -6$. Equation of your curve is $y= x^3 - 6x^2 + 9x - 6$.
 January 8th, 2014, 11:57 PM #3 Senior Member   Joined: Dec 2012 Posts: 372 Thanks: 2 Re: Curve with gradient function $y'(x)= 3x^2 - 12x +9 \ \Rightarrow \ y#39;(2) = -3$. This is the gradient at M. To find the equation of the normal to the curve at M, we first note that the normal vector must be perpendicular to the tangent vector. Since the tangent has slope of -3, the normal must have a slope of $\frac{1}{3}.$ The normal has equation $\dfrac{y + 4}{x - 2}= \dfrac{1}{3}$. $y(x)= \int (3x^2 - 12x +9)dx = x^3 - 6x^2 + 9x + k$ $y(2)= -4 \ \Rightarrow \ 8 - 24 + 18 + k = - 4 \ \Rightarrow \ k = -6$. Equation of your curve is $y= x^3 - 6x^2 + 9x - 6$.
 January 9th, 2014, 02:27 AM #4 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 Re: Curve with gradient function Thanks . I can understand it with your solution .

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post jiasyuen Calculus 1 January 9th, 2014 12:51 PM jiasyuen Calculus 5 January 9th, 2014 10:55 AM BenR Calculus 0 December 12th, 2012 01:33 AM Wooseok Nam Applied Math 0 May 20th, 2010 12:44 AM Mediocre1981 Calculus 3 August 29th, 2008 03:01 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top