1. Home
  2. Calculators
  3. Calculators: Calculus III
  4. Calculus Calculator

Tangent Plane Calculator

Find tangent planes step by step

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

((
,
,
))

If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please contact us.

Your Input

Calculate the tangent plane to x2+y2+z2=14x^{2} + y^{2} + z^{2} = 14 at (x,y,z)=(1,3,2)\left(x, y, z\right) = \left(1, 3, 2\right).

Solution

The function can be represented in the form F(x,y,z)=0F{\left(x,y,z \right)} = 0, where F(x,y,z)=x2+y2+z2−14F{\left(x,y,z \right)} = x^{2} + y^{2} + z^{2} - 14.

Find the partial derivatives.

∂∂x(F(x,y,z))=∂∂x(x2+y2+z2−14)=2x\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).

∂∂y(F(x,y,z))=∂∂y(x2+y2+z2−14)=2y\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).

∂∂z(F(x,y,z))=∂∂z(x2+y2+z2−14)=2z\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.

∂∂x(x2+y2+z2−14)∣((x,y,z)=(1,3,2))=(2x)∣((x,y,z)=(1,3,2))=2\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

∂∂y(x2+y2+z2−14)∣((x,y,z)=(1,3,2))=(2y)∣((x,y,z)=(1,3,2))=6\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

∂∂z(x2+y2+z2−14)∣((x,y,z)=(1,3,2))=(2z)∣((x,y,z)=(1,3,2))=4\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 ∂∂x(F(x,y,z))∣((x,y,z)=(x0,y0,z0))(x−x0)+∂∂y(F(x,y,z))∣((x,y,z)=(x0,y0,z0))(y−y0)+∂∂z(F(x,y,z))∣((x,y,z)=(x0,y0,z0))(z−z0)=0.\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(x−1)+6(y−3)+4(z−2)=02 \left(x - 1\right) + 6 \left(y - 3\right) + 4 \left(z - 2\right) = 0.

This can be rewritten as 2x+6y+4z=282 x + 6 y + 4 z = 28.

Or, more simply: z=−x2−3y2+7z = - \frac{x}{2} - \frac{3 y}{2} + 7.

Answer

The equation of the tangent plane is z=−x2−3y2+7=−0.5x−1.5y+7z = - \frac{x}{2} - \frac{3 y}{2} + 7 = - 0.5 x - 1.5 y + 7A.