September 30th, 2019, 07:47 PM  #1 
Newbie Joined: Sep 2019 From: Papua New Guinea Posts: 8 Thanks: 1  help
Find the slope of the parabola y=x^2 at the point P(3,4), and use it to find the equation of the tangent line to y=x^ at x=3. 
September 30th, 2019, 08:23 PM  #2 
Senior Member Joined: Sep 2016 From: USA Posts: 670 Thanks: 440 Math Focus: Dynamical systems, analytic function theory, numerics 
Ok I did it. Now tell me your answer so I can check my work.

September 30th, 2019, 09:45 PM  #3  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,304 Thanks: 961 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Dan  
October 1st, 2019, 04:27 PM  #4 
Senior Member Joined: Sep 2016 From: USA Posts: 670 Thanks: 440 Math Focus: Dynamical systems, analytic function theory, numerics  
October 1st, 2019, 07:22 PM  #5 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,304 Thanks: 961 Math Focus: Wibbly wobbly timeywimey stuff. 
Hmmmm... I wonder if there's a metric that will do the trick so that the point really is on the parabola. Have to think that one over. Dan 
October 1st, 2019, 11:44 PM  #6 
Senior Member Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 132 Thanks: 49 Math Focus: Area of Circle 
it is probably P(3,9).

October 2nd, 2019, 02:40 AM  #7 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,304 Thanks: 961 Math Focus: Wibbly wobbly timeywimey stuff.  
October 2nd, 2019, 02:43 AM  #8  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,304 Thanks: 961 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
The slope of the line tangent to the curve at (3, 9) (or whatever) is the value of the first derivative at that point. So $\displaystyle y = x^2 \implies \dfrac{dy}{dx} = 2x$. That gives you the slope of a line (the derivative) and a point on that line (3, 9). How can you find the equation of the line from here? Dan  
October 2nd, 2019, 03:00 AM  #9 
Senior Member Joined: Feb 2010 Posts: 714 Thanks: 151 
Perhaps the coordinates were correct but the OP mistakenly thought it was on the curve. Then the question is "find the equation of the tangent line to $\displaystyle y=x^2$ that goes through $\displaystyle (3,4)$." If that was the question then I get two solutions: $\displaystyle y=(6 \pm 2\sqrt{5})(x3)+4$ 
October 2nd, 2019, 04:20 AM  #10  
Senior Member Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 132 Thanks: 49 Math Focus: Area of Circle  Quote:
$y'=2x$ therefore, slope of g(x) is $2x_1$. $2{x_1}^2+b={x_1}^2 \Rightarrow b={x_1}^2$. $g(x)=2{x_1}x{x_1}^2$ $g(3)=6{x_1}{x_1}^2=4 \Rightarrow x_1=3 \pm \sqrt{5}$ $g(x)=(3 \pm \sqrt{5})x  (3 \pm \sqrt{5})^2$ Which means there is no tangent that passes through $(3,4)$.  