Intersection of the Slopes of Curves

**StarlitxSunshine** · November 7th 2009, 02:14 PM

If m is the slope of the curve $xy=3$ and m2 is the slope of the curve $x^2-y^2=4$ , then at a point of intersection of the two curves:

(A) m1 = m2
(B) m1= -m2
(C) m1*m2 = 1
(D) m1*m2 = -1
(E) m1*m2 = -2

I think I might be doing it wrong... >_<

First, I found the derivative of the first one explicitly.
$ \frac{dy}{dx}[3x^-1] $

which is:

$ \frac{-3}{x^2} $

And then, the derivative of the second one implicitly:

$ \frac{dy}{dx}[x^2-y^2]=\frac{dy}{dx} [4]$

which is:

$ \frac{dy}{dx}= \frac{x}{y}$

But I can't see a relationship between the two slopes...?

**tonio** · November 7th 2009, 02:25 PM

Originally Posted by StarlitxSunshine

If m is the slope of the curve $xy=3$ and m2 is the slope of the curve $x^2-y^2=4$ , then at a point of intersection of the two curves:

(A) m1 = m2
(B) m1= -m2
(C) m1*m2 = 1
(D) m1*m2 = -1
(E) m1*m2 = -2

I think I might be doing it wrong... >_<

First, I found the derivative of the first one explicitly.
$ \frac{dy}{dx}[3x^-1] $

which is:

$ \frac{-3}{x^2} $

And then, the derivative of the second one implicitly:

$ \frac{dy}{dx}[x^2-y^2]=\frac{dy}{dx} [4]$

which is:

$ \frac{dy}{dx}= \frac{x}{y}$

But I can't see a relationship between the two slopes...?

Multiply both slopes...

Tonio

**StarlitxSunshine** · November 7th 2009, 02:30 PM

O.o I tried that...

$ \frac{-3}{x^2} * \frac{x}{y} = \frac{-3}{xy} $

That doesn't help... ? -3 isn't a choice in the answers...

**Debsta** · November 7th 2009, 02:31 PM

You need to consider the slopes "at the point of intersection". So you need to find that point first, evaluate both gradients (at that point) then look for the answer.

**StarlitxSunshine** · November 7th 2009, 02:33 PM

Originally Posted by Debsta

You need to consider the slopes "at the point of intersection". So you need to find that point first, evaluate both gradients (at that point) then look for the answer.

Ohh...Umm.. :33

To find the point of intersection, should I make the table of values for each equation ? Or is there another way ?

**Debsta** · November 7th 2009, 02:35 PM

Sub y = 3/x from first eqtn into the second one and do it algebraically.

**tonio** · November 7th 2009, 02:38 PM

Originally Posted by StarlitxSunshine

O.o I tried that...

$ \frac{-3}{x^2} * \frac{x}{y} = \frac{-3}{xy} $

That doesn't help... ? -3 isn't a choice in the answers...

But you're given what xy is, girl!

Tonio

**tonio** · November 7th 2009, 02:39 PM

Originally Posted by StarlitxSunshine

Ohh...Umm.. :33

To find the point of intersection, should I make the table of values for each equation ? Or is there another way ?

You don't need to find the intersection point at all: it is a very nasty thing (square roots inside square roots...you don't want to meet such a thing in an obscure alley at night!)!

Tonio

**StarlitxSunshine** · November 7th 2009, 02:43 PM

Originally Posted by tonio

But you're given what xy is, girl!

Tonio

O_O I am!!

I'm so sorry !! X3

Then it would be -3/3 = -1 which is a choice @_@

Sorry, sorry !

**Debsta** · November 7th 2009, 02:45 PM

That's true. Tonio's way is much easier. You're nearly there.

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