# Gradient Problem: Tangent plane of a surface

So I'm stuck on the following problem:

Write the equation of the plane tangent to

$z = f(x,y) = (x^2+y^3-8)^{1\over4}$ at $(5,4,3)$

I know how to find the equation of the plane but usually the $z$ is part of the function. I was wondering if all I had to do was subtract the $z$ and set the whole thing to zero or do I do this as is.

I tried that and this is the answer that I got: $${5\over54}\left(x-5\right)+{4\over9}\left(y-4\right)-\left(z-3\right) = 0$$ is that right?

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What's $k{}{}$? –  joriki Nov 26 '12 at 7:49
supposed to be z, i fixed it now –  StealzHelium Nov 26 '12 at 7:53
Your answer is correct. –  Mhenni Benghorbal Nov 26 '12 at 8:18
I assume by "usually $z$ is part of the function" that you mean you are used to finding plane equations tangent to surfaces defined by $F(x,y,z)=c$. If that's true, then what you have done is fine.
Alternatively, when a surface is defined by $z=f(x,y)$, then the vectors $\left\langle1,0,\frac{\partial f}{\partial x}(x_0,y_0)\right\rangle$ and $\left\langle0,1,\frac{\partial f}{\partial y}(x_0,y_0)\right\rangle$ will be parallel to the plane at $(x_0,y_0,f(x_0,y_0))$. So you can use them to give a parametric description of the plane, or cross them to find a normal vector for the plane. If you do cross them, you find that $\left\langle\frac{\partial f}{\partial x}(x_0,y_0),\frac{\partial f}{\partial y}(x_0,y_0),-1\right\rangle$ is a normal vector.
which of the two approaches do you think this problem is going for? I should also note that its part a, where part b asks to use tangent-plane approximation to estimate $(5.0216^2+3.9973^3-8)^{1/4}$ –  StealzHelium Nov 26 '12 at 8:10
I don't think it would matter what approach you take, seeing as how they both answer the question that is asked of you. If you think your instructor might prefer to see one over the other, you should talk to them about it. Either approach will give you a tangent plane equation, and then you can substitute in $x=5.0216$ and $y=3.9973$ and solve for $z$ to answer part b. –  alex.jordan Nov 26 '12 at 19:18