My Math Forum help

 Calculus Calculus Math Forum

 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. Thanks from gwen110
 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. Thanks from topsquark
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 timey-wimey stuff.
Quote:
 Originally Posted by nantii fariiman 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.
Quote:
 Originally Posted by SDK Ok I did it. Now tell me your answer so I can check my work.
Well I couldn't do it! (3, 4) isn't even a point on the parabola!

-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
Quote:
 Originally Posted by topsquark Well I couldn't do it! (3, 4) isn't even a point on the parabola! -Dan
lol that's what I get for being snarky.

 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 timey-wimey 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). Thanks from topsquark
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 timey-wimey stuff.
Quote:
 Originally Posted by tahirimanov19 it is probably P(3,9).
Nah! That would be too easy.

-Dan

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 timey-wimey stuff.
Quote:
 Originally Posted by nantii fariiman 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.
Okay, back to the actual question.

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: $\displaystyle y=(6 \pm 2\sqrt{5})(x-3)+4$ Thanks from topsquark and tahirimanov19
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:
 Originally Posted by mrtwhs find the equation of the tangent line to $\displaystyle y=x^2$ that goes through $\displaystyle (3,4)$
The function of slop would be $g(x)=ax+b$. Let the tangent point be $(x_1, y_1)$. Therefore, $y_1=ax_1+b={x_1}^2$.
$y'=2x$ 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 $(3,4)$.

 Thread Tools Display Modes Linear Mode

 Contact - Home - Forums - Cryptocurrency Forum - Top