Finished your HSC? Check out our University Discussion, Non-School forums or help out next year's HSC students!
YOU can help the next generation of students in the community! Share your trial papers and notes on our Notes & Resources page

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

You must log in or register to reply here.

Share:

Facebook Twitter Reddit Pinterest Tumblr WhatsApp Email Link

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top