Equation of a normal?

Question

I would just like to check that I have done this correctly. I have to find the equation of the normal to the curve 3x^2+2xy^2+6=2y^3 at the point (2,3).

First I have differentiated wrt x giving: dy/dx = 3x+y^2/3y^2-2xy.

Then using x=2 and y =3: dy/dx = 1.

So the equation of the tangent is (using y=mx+c): 3=(1)(2)+c.

So c=1. Therefore y=x+1.

Now the normal to this is: y=(-1)x+c, passing through (2,3).

So 3=(-1)(2)+c, so c=5. Therefore the equation of the normal is y=-x+5.

Is this correct.

Answer 1

here's how i would do this one;
3x^2+2xy^2+6=2y^3
differentiate,
6x+4xydy/dx+2y^2=6y^2dy/dx
substitute point (2,3)
12+24dy/dx+18=54dy/dx
30dy/dx=30
dy/dx=1 at(2,3)

equation to normal of curve;
y-y1= (-1/dy/dx(2,3))(x-x1)
(x1,y1)=(2,3),-1/dy/dx= -1
hence,
y-3= -1(x-2)
whence,
y+x-5=0,as required

this is another method to
back up your method which
was spot on!

i hope that this helps

Answer 2

Yes it's correct. You did it right.

But you could have gone through a slightly shorter process. The differentiation part is standard, can't get away from it in this problem. However, as for looking for the equation of the normal, you could have gone through a shorter route.

On replacing the values of x and y in the value you got for dy/dx, you actually obtain the value of the gradient of the tangent to the curve at the point (2,3).

Then, you looked for the equation of tangent...etc

However, you could have done it easily by finding the gradient of the normal directly from the gradient of the tangent.

Let's say that the gradient of a tangent at a point is m
Then, the gradient of a normal at the same point
= -( 1/ m )
Hence, if your tangent's gradient is say 2, the normal's gradient is -1/2

If you apply this in the current problem, then normal's gradient will be -1/1 which gives you -1.

It's easier and saves time and space. Hope this helps.

Answer 3

I followed all of your steps and worked it out myself and I obtained the same results as you. I do not detect any inaccuracies or flaws in reasoning.

Just one side-note: when finding the equation of the tangent like, I notice you use slope-intercept form. I don't personally generally do this, though in the above example it seems to work just fine. Instead I use point-slope form, which states that, given a point P(x1, y1) and slope M, the equation of the line is:

y - y1 = M(x - x1)

This seems like a more direct approach, since you have all of the variables in hand. The slope-intercept form is asking for something that you are instead deducing. I have only recently gotten comfortable with this form myself. I don't remember it at all from high school algebra, but they ingrained point-slope into my deepest brain cells.... lol

--charlie

Answer 4

No!
Please give me best answer thanks!