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Let $C$ be the curve through $P\left(-2,1,5\right)$ intersecting the surfaces $\Sigma_1:z=x^2+y^2$ and $\Sigma_2:z=13-(x^2+4y^2)$. Find the tangent line of $C$ at $P$.

My teacher used the total derivative of each surface to find the orthogonal vector. The technique is simple, I just don't fully understand why she did so. Can anyone please provide a geometric explanation? Thanks!

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up vote 2 down vote accepted

The total derivative of a function $f:\mathbb{R}^3\rightarrow\mathbb{R}$ at a point $P=\mathbf{x}_0$ is a linear function $df:\mathbb{R}^3\rightarrow\mathbb{R}$, $$ df =\frac{\partial f}{\partial x}dx +\frac{\partial f}{\partial y}dy +\frac{\partial f}{\partial z}dz, $$ i.e. so that $df(\Delta x,\Delta y,\Delta z) =\frac{\partial f}{\partial x}\Delta x +\frac{\partial f}{\partial y}\Delta y +\frac{\partial f}{\partial z}\Delta z$. It gives you the linear approximation $$ f(\mathbf{x}) \approx f(\mathbf{x}_0) +\frac{\partial f(\mathbf{x}_0)}{\partial x}\Delta x +\frac{\partial f(\mathbf{x}_0)}{\partial y}\Delta y +\frac{\partial f(\mathbf{x}_0)}{\partial z}\Delta z $$ where $(x,y,z)=\mathbf{x}=\mathbf{x}_0+\Delta\mathbf{x}$ and $\Delta\mathbf{x}=(x-x_0,y-y_0,z-z_0)$, and its level sets are planes in $\mathbb{R}^3$ perpendicular to the gradient $\nabla{f}(\mathbf{x}_0)$.

In our case, if we take the cross product of the gradients at a point of common intersection $\mathbf{x}_0$, we will also find the tangent to the curve (not line!) $C$, since this gives us something normal to both gradients and therefore tangent to $\Sigma_1$ and $\Sigma_2$.

Now $\Sigma_1$ is a paraboloid with global minimum at the origin and axis of symmetry/revolution along the $z$ axis, and $\Sigma_2$ is similar but inverted, with parabolic vertical cross sections and elliptic horizontal cross sections and global maximum at $(0,0,13)$. To calculate these gradients, we need to write the equations for these curves as functions: $$f_1=x^2+ y^2-z\quad\implies\quad\nabla{f_1}=(2x,2y,-1)$$ $$f_2=x^2+4y^2+z\quad\implies\quad\nabla{f_2}=(2x,8y,+1)$$ (the curves will then be the level sets for the values $f_1=0$ and $f_2=13$). This would give us a tangent vector parallel to $$\nabla{f_1}\times\nabla{f_2}=(10y,-4x,12xy)$$ and hence, at $P(-2,1,5)$, a parametric tangent line (with parameter $s$ being distance from $P$) of $$\mathbf{x}(s)=(-2,1,5)\pm\frac{s}{\sqrt{185}}(5,-4,12).$$

Note also that curve $C$ is the locus of points $(x,y,z)$ satisfying $x^2+y^2=z=13-(x^2+4y^2)$ $\implies$ $2x^2+5y^2=13$ $\implies$ $$ \left(\frac{x}{\sqrt{13/2}}\right)^2+ \left(\frac{y}{\sqrt{13/5}}\right)^2= 13 $$ which could therefore be parametrized by $$ \mathbf{x}(t)= \left( \sqrt{\frac{13}2}\cos{t}, \sqrt{\frac{13}5}\sin{t}, \frac{13}{10} \left(5\cos^2t+3\sin^2t\right) \right), $$ giving yet another way to compute the tangent!

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