Tangent Plane Calculator

The calculator will try to find the tangent plane to the explicit and the implicit curve at the given point, with steps shown.

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Your Input

Calculate the tangent plane to $$$x^{2} + y^{2} + z^{2} = 14$$$ at $$$\left(x, y, z\right) = \left(1, 3, 2\right)$$$.

Solution

The function can be represented in the form $$$F{\left(x,y,z \right)} = 0$$$, where $$$F{\left(x,y,z \right)} = x^{2} + y^{2} + z^{2} - 14$$$.

Find the partial derivatives.

$$$\frac{\partial}{\partial x} \left(F{\left(x,y,z \right)}\right) = \frac{\partial}{\partial x} \left(x^{2} + y^{2} + z^{2} - 14\right) = 2 x$$$ (for steps, see partial derivative calculator).

$$$\frac{\partial}{\partial y} \left(F{\left(x,y,z \right)}\right) = \frac{\partial}{\partial y} \left(x^{2} + y^{2} + z^{2} - 14\right) = 2 y$$$ (for steps, see partial derivative calculator).

$$$\frac{\partial}{\partial z} \left(F{\left(x,y,z \right)}\right) = \frac{\partial}{\partial z} \left(x^{2} + y^{2} + z^{2} - 14\right) = 2 z$$$ (for steps, see partial derivative calculator).

Evaluate the derivatives at the given point.

$$$\frac{\partial}{\partial x} \left(x^{2} + y^{2} + z^{2} - 14\right)|_{\left(\left(x, y, z\right) = \left(1, 3, 2\right)\right)} = \left(2 x\right)|_{\left(\left(x, y, z\right) = \left(1, 3, 2\right)\right)} = 2$$$

$$$\frac{\partial}{\partial y} \left(x^{2} + y^{2} + z^{2} - 14\right)|_{\left(\left(x, y, z\right) = \left(1, 3, 2\right)\right)} = \left(2 y\right)|_{\left(\left(x, y, z\right) = \left(1, 3, 2\right)\right)} = 6$$$

$$$\frac{\partial}{\partial z} \left(x^{2} + y^{2} + z^{2} - 14\right)|_{\left(\left(x, y, z\right) = \left(1, 3, 2\right)\right)} = \left(2 z\right)|_{\left(\left(x, y, z\right) = \left(1, 3, 2\right)\right)} = 4$$$

The equation of the tangent plane is $$$\frac{\partial}{\partial x} \left(F{\left(x,y,z \right)}\right)|_{\left(\left(x, y, z\right) = \left(x_{0}, y_{0}, z_{0}\right)\right)} \left(x - x_{0}\right) + \frac{\partial}{\partial y} \left(F{\left(x,y,z \right)}\right)|_{\left(\left(x, y, z\right) = \left(x_{0}, y_{0}, z_{0}\right)\right)} \left(y - y_{0}\right) + \frac{\partial}{\partial z} \left(F{\left(x,y,z \right)}\right)|_{\left(\left(x, y, z\right) = \left(x_{0}, y_{0}, z_{0}\right)\right)} \left(z - z_{0}\right) = 0.$$$

In our case, $$$2 \left(x - 1\right) + 6 \left(y - 3\right) + 4 \left(z - 2\right) = 0$$$.

This can be rewritten as $$$2 x + 6 y + 4 z = 28$$$.

Or, more simply: $$$z = - \frac{x + 3 y - 14}{2}$$$.

Answer

The equation of the tangent plane is $$$z = - \frac{x + 3 y - 14}{2} = - 0.5 x - 1.5 y + 7$$$A.