Question

At the given point, find the slope of the curve, the line that is tangent to the curve, or the line that is normal to the curve

Answer 1

y^4 +x^3 = y^2 + 9x, tangent at (0,1)

Answer 2

Oo, this is interesting.

Answer 3

First we verify that the point lies on the curve: 1 + 0 = 1 + 0 is true.

Answer 4

You know one point on the tangent. To know the tangent, you need to also know its slope. To find the slope, differentiate both sides of the equation (this is called implicit differentiation) and solve for y'.

Answer 5

Doing so you should get:
\[\begin{align*}
y^4 + x^3 &= y^2 + 9x\\
4y^3y' + 3x^2 &= 2yy' + 9\\
y' &= \frac{9-3x^2}{4y^3-2y} = \frac{9-3\cdot0^2}{4\cdot1^3-2\cdot1} = \frac{9}{2}
\end{align*}\]
And that's the slope.

Answer 6

Any questions so far?

Answer 7

No, I'm understanding it so far.

Answer 8

Can you use the two facts (the point and the slope) to find the line on your own?

Answer 9

I got \[\frac{ 9 }{ 2 }x + 1\]

I think that's right.

Answer 10

yes

Answer 11

Would you like to see another solution?

Answer 12

@robtobey Would you like to see it?

Answer 13

Plots and a solution using Mathematica 9 Home Edition is attached.

Answer 14

Thanks!

Answer 15

Gather all terms on one side of the equation:
\[-y^2+y^4-9x+x^3 = 0\]
Interpret the left-hand side as a polynomial:
\[p = -y^2+y^4-9x+x^3\]
Translate the polynomial so that \((0, 1)\) gets moved to the origin:
\[\begin{split}
p(x,\ y+1) &= -(y+1)^2 + (y+1)^4 - 9x + x^3\\
&= -1-2y-y^2\ +\ 1+4y+6y^2+4y^3+y^4\ -\ 9x\ +\ x^3\\
&= 2y + 5y^2 + 4y^3 + y^4 - 9x + x^3
\end{split}\]
Translate it back.
\[\begin{split}
p &= p(x,\ y+1)(x,\ y-1)\\
&= 2(y-1) + 5(y-1)^2 + 4(y-1)^3 + (y-1)^4 - 9x + x^3
\end{split}\]
We are interested in how the polynomial behaves near \((0, 1)\). Near that point, \((y-1)\) and \(x\) are small. If we truncate all higher order terms (higher than \(1\) in \((y-1)\) and \(x\)), we will get a polynomial which behaves similarly to \(p\) near \((0, 1)\)
\[\begin{split}
q &= 2(y-1) - 9x
\end{split}\]
Set \(q\) to zero and you'll get the equation of the tangent line.

Answer 16

Note how this approach "saved" you from having to take derivatives.

Answer 17

A plot showing the normal line, that was omitted from the first posting, as well as the tangent line and the first quadrant curve is attached.

Answer 18

@JoelSjogren
I want to be "saved", however, I've run out of time.
Will get back here later.

Thank you for your effort.

Answer 19

No problem both of you :)