The function is as follows: $$\begin{bmatrix}u \\ v \end{bmatrix}= f(x,y,z)= \begin{bmatrix}e^{x-y}+y \\ \sin(x^2-z) \end{bmatrix}$$ and I need to obtain the equation of the tangent plane. The equations I got are $u=x+1$ and $v=2x-z-1$. The answer I got is wrong according to the key but I am not able to see the mistake. Where have I gone wrong? ... Added: Sorry, I forgot to put in the point $(x,y,z,u,v)=(1,1,1,2,0)$ on which the tangent plane should be calculated.
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Would you care to show us how you got your equations? – M.B. Nov 19 '13 at 14:26
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Sure :) First I found the respective partial derivatives along x, y and z. Then I substituted for the point (which I added later... sorry I skipped that). Finally, I solved the following equation:$$\begin{bmatrix}u-2\v-0\end{bmatrix} = f_x(x-1)+f_y(y-1)+f_z(z-1)$$ – Artemisia Nov 19 '13 at 14:41
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What's your function? Is it a curve or a surface? – Shuchang Nov 20 '13 at 08:20
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They haven't mentioned it :) – Artemisia Nov 20 '13 at 08:21
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I am not sure I understand what you have meant. Do you mean to say $u$ and $v$ are each functions of $x,y,z$, therefore $u = f_1(x,y,z) = \exp(x-y)+y$ and $v = f_2(x,y,z) = \sin(x^2 -z)$? – Mark Fantini Nov 25 '13 at 01:17