The slope of the tangent line to a curve at any point (x, y) on the curve is x divided by y. What is the equation of the curve if (4, 1) is a point on the curve?

Question

jhannybean · Answer

So does this mean $$\frac{dy}{dx} = \frac{x}{y}$$

jhannybean · Answer

If so, $$ydy = xdx$$Im kind of new to this.... just attempting.

jhannybean · Answer

$$\int ydy=\int xdx$$$$\frac{1}{2}y^2=x^2+C$$$$y=\sqrt{2(x^2+C)}~? $$

anonymous · Answer

the answer only give them as y^2 and x^2

jhannybean · Answer

I havent taken ordinary differentials yet, so i'm just attempting, I may be wrong. It looked like a fun problem to try.

amistre64 · Answer

1/2 x^2 + C

anonymous · Answer

no wait sorry i was wrong

jhannybean · Answer

Ah ok.

irishboy123 · Answer

there's a mistake in the integration -- look at it again......

jhannybean · Answer

$$y^2=\frac{1}{2}x^2+C$$

amistre64 · Answer

$$\frac12y^2=\frac12x^2+C$$
$$y^2-x^2=C$$

jhannybean · Answer

Oh is that how it works? I see.

anonymous · Answer

they have
 x^2-y^2=15
x^2+y^2=15
x+y=15
xy=15

amistre64 · Answer

who is 'they'?

jhannybean · Answer

So then we can easily find C right? We're given the point (4,1) and we have our equation $$y^2-x^2=C$$$$(1)^2-(4)^2=C$$$$1-16=C$$$$C=-15$$

amistre64 · Answer

(4,1) is read as (x=4, y=1)  but in this case its not really that important

amistre64 · Answer

dbl chk

y^2 - x^2 = -15

2y y' - 2x = 0

y y' - x = 0

y' =x/y