Differentiation Problem

yonski · 2006-10-19 21:30:42

Hi there,
just going through some differentiation exercises but i've found one or two i can't figure out...

1. (a) If (1+x)(2+y) = x^2 + y^2 , find dy/dx in terms of y and x.
(b) Show that there are two points at which the tangents to the curve are parallel to the y-axis.

Now i've done part (a) and found dy/dx = (2x-2-y)/(1+x-2y) . I think that's right, but how on earth do you do part (b)?!

And this is the second question...

2. Find the coordinates of the turning point on the curve y^3 + 3xy^2 - x^3 = 3 .

I thought it looked relatively simple - find dy/dx then set it as equal to zero. But when i do this i end up with
(x^2 - y^2)/(y+2x) = 0 , and i don't know where to go from there lol.

Thanks for any help!
Jon.

Last edited by yonski (2006-10-19 21:34:16)

George,Y · 2006-10-20 03:34:51

dy/dx is tangent slope, so you can find some time when 1+x-2y is zero while 2x-2-y not.

A turning point also satisfies a condition that defferientials have opposite signs between on the left and on the right.

So that the slopes are like this /-\ or like this \-/

x^2-y^2=0 => (x+y)(x-y)=0 => x=y or x=-y

mathsyperson · 2006-10-20 03:41:43

I've just worked through 1a) and I get the same answer as you, which is good.

For 1b), it wants the points at which the tangent is parallel to the y-axis. This means that the gradient would have to be undefined, and so the denominator of dy/dx would have to be 0.

So now we have 1+x-2y = 0, or y = (x+1)/2. However, we need to find the points at which the original equation works as well.

So now we have (1+x)(2+(x+1)/2) = x² + ((x+1)/2)², by substituting in the value of y.

This looks rather nasty, but by rearranging it for a bit we can turn it into the much nicer 3x²-10x-9 = 0. This is a quadratic equation that comes out as x = (5±2√(13))/3

So that's the x-value of your two points, and then you can get the y-values by substituting the x's into one of your equations.

2) is solved in the same way.
You rightly set (x^2 - y^2)/(y+2x) to 0, and now you just need to find values of x and y that satisfy both that and the original cubic equation.

yonski · 2006-10-20 07:09:06

Thanks, that's cleared those up. I was almost there with em, just required a little more thinking outside the box with the first one. So if the gradient is infinite/undefined, there's likely to be a division by zero in there?

Last edited by yonski (2006-10-20 07:10:32)

George,Y · 2006-10-20 15:19:15

Yes, you've found it out! Infinity here in fact means some 1/0 or a/0, derived by x and y displacement.

Actually the slope is a kind of directional thinking while the real world is not.

Think of this argument.

Tom loves Maria. So they get married.

In fact it should be this: Tom loves Maria, while Maria loves Tom. So they get married.

The y-x slope is just a tool looking at a line's angle by checking some displacement of x and y from y to x. There is, though, another alternative, x-y slope. And this slope should be 0 without any doubt or undefination.

George,Y · 2006-10-20 15:21:20

I mean, how many times is 1 as nothing?

yonski · 2006-10-22 07:34:03

What happens if Tom loves Maria, but Maria doesn't love him back?

Last edited by yonski (2006-10-22 07:34:25)

Math Is Fun Forum

#1 2006-10-19 21:30:42

Differentiation Problem

#2 2006-10-20 03:34:51

Re: Differentiation Problem

#3 2006-10-20 03:41:43

Re: Differentiation Problem

#4 2006-10-20 07:09:06

Re: Differentiation Problem

#5 2006-10-20 15:19:15

Re: Differentiation Problem

#6 2006-10-20 15:21:20

Re: Differentiation Problem

#7 2006-10-22 07:34:03

Re: Differentiation Problem

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