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How do I get the exact gradient, with rational denominator, of the normal of parabola (1 Viewer)

Thread starter Jmmalic220
Start date Oct 20, 2016

J

Jmmalic220

New Member

#1

Hi.
So, I need to get the exact gradient of the normal of the parabola y^2=12x at the point where x=4 in the first quadrant.
I get near the answer which is -(2√3)/3 but I couldn't get the working right.

Any help is very much appreciated.
Thank you.

E

Esse

xD

#2

Re: How do I get the exact gradient, with rational denominator, of the normal of para

y^2 = 12x
y = root(12x)
y' = 6/root(12x)
when x = 4, tangent gradient = 3/2root3
normal gradient = negative reciprocal of tangent gradient = -2root3/3

M

Mahan1

Member

#3

Re: How do I get the exact gradient, with rational denominator, of the normal of para

Jmmalic220 said:
Hi.
So, I need to get the exact gradient of the normal of the parabola y^2=12x at the point where x=4 in the first quadrant.
I get near the answer which is -(2√3)/3 but I couldn't get the working right.

Any help is very much appreciated.
Thank you.

This is slightly more general method:
let's deferential with respect to x:
$\frac{d}{dx}(y^{2} ) = \frac{d}{dx}(12x)$

$2y \frac{d}{dx}y = 12 \ \ \textrm{(using chain rule) that implies }$

$\textrm{at} \ \ x=4, y= 4 \sqrt{3}$

$\frac{dy}{dx} = \frac{6}{4 \sqrt{3}} = \frac{3}{2 \sqrt{3}}$

$\textrm{since tangent line is perpendicular to the normal at that point, then the slope of the normal is: }$

$- \frac{2 \sqrt{3}}{3}$

Last edited: Nov 27, 2016

pikachu975

Premium Member

#4

Re: How do I get the exact gradient, with rational denominator, of the normal of para

Mahan1 said:
This is slightly more general method:
let's deferential with respect to x:
$\frac{d}{dx}(y^{2} ) = \frac{d}{dx}(12x)$

$2y \frac{d}{dx}y = 12 \ \ \textrm{(using chain rule) that implies }$

$\textrm{at} \ \ x=4, y= 4 \sqrt{3}$

$\frac{dy}{dx} = \frac{6}{4 \sqrt{3}} = \frac{3}{2 \sqrt{3}}$

$\textrm{since tangent line is perpendicular to the normal at that point, then the slope of the normal is: }$

$- \frac{2 \sqrt{3}}{3}$

Is this even 2 unit?

Squar3root

realest nigga

#5

Re: How do I get the exact gradient, with rational denominator, of the normal of para

pikachu975 said:
Is this even 2 unit?

it's not, it's using implicit differentiation learned in 4U. Esse's method is the approach for 2u

S

si2136

Well-Known Member

#6

Re: How do I get the exact gradient, with rational denominator, of the normal of para

Squar3root said:
it's not, it's using implicit differentiation learned in 4U. Esse's method is the approach for 2u

Are you allowed to use 4U methods in 2U?

I remembered implicit in 3U for circle differentiation etc.

I

InteGrand

Well-Known Member

#7

Re: How do I get the exact gradient, with rational denominator, of the normal of para

$\noindent Another way is to just find the `slope' of the normal imagining $y$ to be the first (`independent') variable instead, and then just invert this in the end. So we want the slope of normal to $x =\frac{y^{2}}{12}$ at the point $y = \sqrt{48} = 4\sqrt{3}$, treating $y$ as the first variable for now. Since $\frac{\mathrm{d}x}{\mathrm{d}y} = \frac{2y}{12} = \frac{y}{6}$, the tangent (treating $y$ as first variable) at the desired point has slope $\frac{4\sqrt{3}}{6} = \frac{2\sqrt{3}}{3}$. We need to negate and invert this to get the slope of the normal, but we'll invert it back to get the slope when treating $x$ as the first variable as is usual. So we only need negate this to get the final answer. Thus the answer is $-\frac{2\sqrt{3}}{3}$.$

Last edited: Nov 28, 2016

pikachu975

Premium Member

#8

Re: How do I get the exact gradient, with rational denominator, of the normal of para

si2136 said:
Are you allowed to use 4U methods in 2U?

I remembered implicit in 3U for circle differentiation etc.

There's no 4U methods to use in 2U pretty much... and you won't even need to do the 2U HSC

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