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Multivariable Equation of the Tangent Plane to a Surface at a Point (Example 1)
In single variable calculus finding the equation of a tangent is easy.
But, in multivariable calculus, you’ll be asked to find the equation of a tangent plane to a surface for a given point.
Let’s do a practice example.
Find the equation of the tangent plane to the given surface at the given point, where the surface is \(z = 3{y^2} – 2{x^2} + x\) and the given point is (2,-1,3).
The equation for tangent plane is given by:
\[z – {z_O} = \frac{{\partial z(2, – 1)}}{{\partial x}}(x – {x_O}) + \frac{{\partial z(2, – 1)}}{{\partial y}}(y – {y_O})\]
First find the partial derivatives and substitute in the given point:
\[\frac{{\partial z}}{{\partial x}} = – 4x + 1\]
\[\frac{{\partial z(2, – 1)}}{{\partial x}} = – 4(2) + 1\]
\[\frac{{\partial z(2, – 1)}}{{\partial x}} = – 7\]
\[\frac{{\partial z}}{{\partial y}} = 6y\]
\[\frac{{\partial z(2, – 1)}}{{\partial y}} = 6( – 1)\]
\[\frac{{\partial z(2, – 1)}}{{\partial y}} = – 6\]
And use \({z_O}\) from the initial point, as -3.
Substitute all of these into the tangent plane equation.
\[z + 3 = – 7(x – 2) + ( – 6)(y + 1)\]
\[z + 3 = – 7x + 14 – 6y – 6\]
\[z = – 7x – 6y + 5\]
Click here to see an example on finding a unit tangent vector!