2.5 Tangent Planes To Parametric Surfaces
Let \(S\) be the parametric surface traced out by \(\mathbf r(u,v) = x(u,v)\, \mathbf i + y(u,v)\,\mathbf j + z(u,v)\,\mathbf k\), and \(P_0\) the point with position vector \(\mathbf r(u_0,v_0)\). Holding \(u = u_0\) fixed, \(\mathbf r(u_0,v)\) is a vector function of the single parameter \(v\), and defines a curve \(C_1\) lying on \(S\). Its tangent vector at \(P_0\) is \[\mathbf r_v = \frac {\partial x}{\partial v}(u_0,v_0)\,\mathbf i + \frac {\partial y}{\partial v}(u_0,v_0)\,\mathbf j +\frac {\partial z}{\partial v}(u_0,v_0)\,\mathbf k .\]
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Similarly, holding \(v = v_0\) fixed gives a curve \(C_2\), given by \(\mathbf r(u,v_0)\), lying on \(S\), with tangent vector \[\mathbf r_u = \frac {\partial x}{\partial u}(u_0,v_0)\,\mathbf i + \frac {\partial y}{\partial u}(u_0,v_0)\,\mathbf j +\frac {\partial z}{\partial u}(u_0,v_0)\,\mathbf k .\] If \(\mathbf r_u \times \mathbf r_v \neq \mathbf 0\) at \((u_0,v_0)\), the surface is smooth there. For a smooth surface the tangent plane at \(P_0\) is the plane through \(P_0\) that contains the tangent vectors \(\mathbf r_u\) and \(\mathbf r_v\), and \(\mathbf r_u \times \mathbf r_v\) is a normal vector to it.
Example 2.5.1. Find the tangent plane to the surface with parametric equations \[x =u^2 ,\qquad y = v^2 , \qquad z = u + 2v\] at the point \((1,1,3)\).
Solution. With \(\mathbf r(u,v) = u^2 \,\mathbf i + v^2\,\mathbf j + (u + 2v)\,\mathbf k\), \[\mathbf r_u = 2u\,\mathbf i + \mathbf k ,\qquad \mathbf r_v = 2v\,\mathbf j+ 2\,\mathbf k ,\] and a normal vector is \[\mathbf r_u \times \mathbf r_v = \begin {vmatrix} \mathbf i & \mathbf j & \mathbf k\\ 2u & 0 & 1\\ 0 & 2v & 2 \end {vmatrix} = -2v\,\mathbf i -4u\,\mathbf j + 4uv\,\mathbf k .\] The point \((1,1,3)\) corresponds to \(u = 1\), \(v = 1\), where the normal is \(\langle -2, -4, 4\rangle \). So the tangent plane is \[-2(x-1) -4(y-1) + 4(z-3) =0,\qquad \text {that is}\qquad x+ 2y -2z + 3 = 0 .\]
Example 2.5.2. Find the tangent plane to the sphere \(\mathbf r(\phi ,\theta ) = \langle 2\sin \phi \cos \theta , 2\sin \phi \sin \theta , 2\cos \phi \rangle \) at the point with \(\phi = \frac {\pi }{4}\), \(\theta = 0\).
Solution. The point is \((\sqrt 2, 0, \sqrt 2)\). Differentiating, \begin {align*} \mathbf r_\phi & = \langle 2\cos \phi \cos \theta ,\ 2\cos \phi \sin \theta ,\ -2\sin \phi \rangle = \langle \sqrt 2, 0, -\sqrt 2\rangle ,\\ \mathbf r_\theta & = \langle -2\sin \phi \sin \theta ,\ 2\sin \phi \cos \theta ,\ 0\rangle = \langle 0, \sqrt 2, 0\rangle . \end {align*}
Then \(\mathbf r_\phi \times \mathbf r_\theta = \langle 2, 0, 2\rangle \), parallel to the position vector of the point, as it must be for a sphere. The tangent plane is \(2(x - \sqrt 2) + 2(z - \sqrt 2) = 0\), or \(x + z = 2\sqrt 2\).
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