+0  
 
0
131
2
avatar

How to do this quickly?

The slow way: f(a) - (x-a)/f'(a)

 

Find the equation of the normal to y= x2 at the point (3, 9)

 

What is the fastest way to do this? The one where the answer you get would be x + 6y = 57

The slow way answer is y = 9.5 - x/6

 

Same answer but I belive the fast one is, well, faster. But I forgot how to do it.

Guest Aug 10, 2017
Sort: 

2+0 Answers

 #1
avatar
+1

What is that variable a in f(a)?

I do it like that and think it is quite fast:

 

y=x2

y'= 2x

y' at (3,9) = 6

slope of normal at (3,9) = -1/6

Thus required equation is x+6y = 3 + 6(9)

x+6y=57

 

idk if this is fast or not actually...

Guest Aug 10, 2017
edited by Guest  Aug 10, 2017
 #2
avatar+78577 
+1

 

For the function x^2, the slope of the tangent line to the curve at any point on the curve is 2x

 

A line perpendicuar to this tangent line will have the slope of  - 1 / [2x ]

 

So....at (3,9)....the slope of a perpendicular line will be    -1 / [ 2 (3) ]  =  -1/6

 

So...the equation of a normal line at (3,9) will be

 

y  = -(1/6) ( x - 3) + 9    =   (-1/6)x + 1/2 + 9    =   (-1/6)x + 19/2

 

In standard form we would have

 

-6y  = x + 57

 

x - 6y + 57  = 0

 

Here's the graph : https://www.desmos.com/calculator/m8bovfpy9w

 

 

cool cool cool

CPhill  Aug 10, 2017

6 Online Users

avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details