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Find a normal vector and a tangent vector to the curve given by the equation: $x^5 + y ^5 =2x^3$ at the point $P(1, 1)$. Find the equation of the tangent line.
Edit: The notes I have:enter image description here

Taking $f(x, y) = x^5 - 2x^3 + y^5 = 0$, I got $m_n = (-1,5), m_t = (5, 1)$ and $x = 5y-4$ for the tangent line.

I'm very much unsure on if this formula is what I should/can use and if I've used it correctly.

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  • $\begingroup$ What problems have you had with this so far? By the way, the tangent line aspect appears to be simple differentiation. $\endgroup$
    – HDE 226868
    Nov 29 '14 at 18:36
  • $\begingroup$ Not sure where to begin with this question, my notes for this topic are very poor, notes refer to solving equations in the form $f(x, y) =c$ any good online explanations for these types of questions greatly appreciated. $\endgroup$ Nov 29 '14 at 18:39
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You can find the slope of the tangent line by differentiating. If you don't want to do implicit differentiation (which may be simpler in this case), you can just do some algebra beforehand: $$x^5+y^5=2x^3 \to y^5=2x^3-x^5 \to y=\sqrt[5]{2x^3-x^5}$$ and differentiate according to the power rule. You have the slope of your tangent line; knowing that it goes through $(1,1)$, you should have enough information to solve for that.

The tangent vector will have a slope exactly the same as that of the tangent line. The normal vector will have a slope that is the negative inverse of that of the tangent vector. If $m_t$ is the slope of the tangent vector, the slope $m_n$ of the normal vector will be $-\frac{1}{m_t}$.

In the method you mentioned, you really just have to find $c$ here to match it up with that equation. You would just re-arrange the equation so that all the constants are on one side of the equation. Here, we find that $$F(x,y)=x^5-2x^3+y^5=c$$ and $c=0$.

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    $\begingroup$ This should be $-2x^3$ right? $\endgroup$ Nov 29 '14 at 19:20
  • $\begingroup$ @JSlocombe95 Yes, you are right. $\endgroup$
    – HDE 226868
    Dec 5 '14 at 1:15
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Differentiate explicitly:

$$5x^4 + 5y^4y' = 6x^2$$ Solve for $y'$

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