The quadratic function - Tangent?? (1 Viewer)

[Michaelmoo]

Ok the question says:

Find the exact gradient, with rational denominator, of the normal to the parabola y^2 = 12x at the point where x = 4 in the first quadrant.

Ok i know y take the square root of both sides to create as function of x sub in 4 etc etc.

But when you sub in x = 4, you only get the gradient for the point in the first quadrant (i.e. 4, root 48)

Just out of curiosity how do you get the gradient thats in the 4th quadrant (i.e. 4, - root 48).

The gradients cant be the same at teh two poitns (draw it).

What do you do to find it out? Do you create a function of y and sub in the corresponding y-value???


Thanks in advance?

 

[Timothy.Siu]

Ok the question says:

Find the exact gradient, with rational denominator, of the normal to the parabola y^2 = 12x at the point where x = 4 in the first quadrant.

Ok i know y take the square root of both sides to create as function of x sub in 4 etc etc.

But when you sub in x = 4, you only get the gradient for the point in the first quadrant (i.e. 4, root 48)

Just out of curiosity how do you get the gradient thats in the 4th quadrant (i.e. 4, - root 48).

The gradients cant be the same at teh two poitns (draw it).

What do you do to find it out? Do you create a function of y and sub in the corresponding y-value???


Thanks in advance?

when u square root both sides, y=+-root(12x), not just y=root(12x)
and u differentiate to find the gradient, not just sub in 4

 

[P.T.F.E]

Gradient of the normal is the reciprical of the gradient of the tangent. Gradient of tangent = y'.

hope that helps u get started

 

[youngminii]

Gradient of the normal is the reciprical of the gradient of the tangent. Gradient of tangent = y'.

hope that helps u get started

Who is this guy?
The gradient of the normal is the INVERSE of the gradient of the tangent. Big difference.

Find the exact gradient, with rational denominator, of the normal to the parabola y^2 = 12x at the point where x = 4 in the first quadrant.
You can always do it this way (just to be a tad neater)
x = y^2 / 12
dx/dy = y/6
dy/dx = 6/y
Subbing y back in (where y is +-sqrt(12x))
dy/dx = 6/(+-sqrt(12x))
When x = 4
dy/dx = 6/(+-sqrt(48))
= 6/(+-4sqrt(3))
= +-3/2sqrt3
For normal, -1/dy/dx
m = -+(2sqrt3)/3
etc.

 

[Michaelmoo]

Who is this guy?
The gradient of the normal is the INVERSE of the gradient of the tangent. Big difference.

Find the exact gradient, with rational denominator, of the normal to the parabola y^2 = 12x at the point where x = 4 in the first quadrant.
You can always do it this way (just to be a tad neater)
x = y^2 / 12
dx/dy = y/6
dy/dx = 6/y
Subbing y back in (where y is +-sqrt(12x))
dy/dx = 6/(+-sqrt(12x))
When x = 4
dy/dx = 6/(+-sqrt(48))
= 6/(+-4sqrt(3))
= 3/2sqrt3
For normal, -1/dy/dx
m = (2sqrt3)/3
etc.

Ok in that line, why did you eliminate the minus part? IS it because in teh 1st quadrant the gradient is poitive and in the second the gradient is negative?

And for the second line isnt it -1/(3/2sqrt3), so it should be -2sqrt3/3???

 

[youngminii]

Ok in that line, why did you eliminate the minus part? IS it because in teh 1st quadrant the gradient is poitive and in the second the gradient is negative?

And for the second line isnt it -1/(3/2sqrt3), so it should be -2sqrt3/3???

Yeah my bad, was being careless
I'll fix it

Edit: I'll just finish it off
Okay so the gradient is either (2sqrt3)/3 or -(2sqrt3)/3
To find the one that's in the first quadrant, let's think geometrically (cbf to prove it)
If it's in the first quadrant, the tangent to the curve is always positive. If it's in the fourth quadrant, the concavity goes the other way and so the tangent is always negative.
The normal is simple the inverse, (ie. swaps y with x) so the normal to any point in the first quadrant will now always be negative, and positive in the fourth quadrant. Hope you understand

This means that when x = 4 in the first quadrant, since it is always negative, the gradient will be -(2sqrt3)/3

Please check for any careless errors =D

 

[tommykins]

y^2 = 12x normal when x = 4

implicit diff

2y(dy/dx) = 12
dy/dx = 6/y = 6/sqrt12x
Now negative recipricol is the normal.
dy/dx = -sqrt12x/6 = -sqrt48/6 = -4sqrt3/6 = -2sqrt3/3

 

[tacogym27101990]

y^2 = 12x normal when x = 4

implicit diff

2y(dy/dx) = 12
dy/dx = 6/y = 6/sqrt12x
Now negative recipricol is the normal.
dy/dx = -sqrt12x/6 = -sqrt48/6 = -4sqrt3/6 = -2sqrt3/3

haha yeah implicit differentiation.
made things so much easier

 

[Michaelmoo]

y^2 = 12x normal when x = 4

implicit diff

2y(dy/dx) = 12
dy/dx = 6/y = 6/sqrt12x
Now negative recipricol is the normal.
dy/dx = -sqrt12x/6 = -sqrt48/6 = -4sqrt3/6 = -2sqrt3/3

Implicit differenciation. Do they teach this at school or a skill you acquire???

 

[tacogym27101990]

its a 4 unit technique but its very easy to pick up and makes these differentiations easier and quicker
i guess you can use it in 2 unit

 

[tommykins]

Implicit differenciation. Do they teach this at school or a skill you acquire???

basically differentiate all the terms.

since we're finding dy/dx - you differentiate y^2 into 2y.dy/dx

so diff y^3 would be 3y^2.dy/dx

differentiate x as normal.

rearrange to find dy/dx

if you're doing 4unit it'll be covered.