Select the variables and enter the function along with its coordinates. The tool will instantly determine the tangent plane to a point on the curve, providing step-by-step calculations.
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The Online Tangent Plane Calculator allows you to compute the equation of a tangent plane to a surface at a specific point quickly and accurately. It supports functions involving two or three variables and delivers step-by-step differentiation results. This tool simplifies complex multivariable calculus calculations while also helping users understand the underlying concepts.
Similar to how the derivative dy/dx gives the tangent line to a single-variable function, partial derivatives determine the tangent plane to a multivariable surface. The tangent plane represents the flat surface that best approximates the curved surface at a given point.

A tangent plane exists only if the function is differentiable at the specified point. Differentiability ensures that partial derivatives are continuous and the surface has a well-defined slope in all directions.
For a surface defined by z = f(x,y) at point P₀ = (x₀, y₀), the tangent plane equation is:
$$ z = f(x_0, y_0) + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) $$
If the surface involves three variables z = f(x,y,z) at P₀ = (x₀, y₀, z₀), the tangent plane becomes:
$$ z = f(x_0, y_0, z_0) + f_x(x_0, y_0, z_0)(x - x_0) + f_y(x_0, y_0, z_0)(y - y_0) + f_z(x_0, y_0, z_0)(z - z_0) $$
Example 1: Find the tangent plane to z = x² + y² at (1,2,5).
Solution:
Partial derivatives:
$$ f_x = 2x, \quad f_y = 2y $$
At (1,2):
$$ f_x = 2, \quad f_y = 4 $$
Tangent plane:
$$ z - 5 = 2(x - 1) + 4(y - 2) $$
Simplified form:
$$ 2x + 4y - z - 5 = 0 $$
Example 2: Tangent plane to f(x,y) = sin(2x)cos(3y) at (π/3, π/4).
Solution:
Partial derivatives:
$$ f_x = 2cos(2x)cos(3y) $$
$$ f_y = -3sin(2x)sin(3y) $$
Substitute the point into the tangent plane formula:
$$ z = f(π/3,π/4) + f_x(π/3,π/4)(x - π/3) + f_y(π/3,π/4)(y - π/4) $$
After simplification:
$$ z = -\frac{\sqrt{6}}{2} + \frac{\sqrt{2}}{2}(x - π/3) - \frac{3\sqrt{6}}{4}(y - π/4) $$
Example 3: Tangent plane to x² + y² + z² = 30 at (1,-2,5).
Solution:
Gradient vector:
$$ \nabla f = (2x, 2y, 2z) $$
At the point:
$$ (2, -4, 10) $$
Tangent plane equation:
$$ 2(x - 1) - 4(y + 2) + 10(z - 5) = 0 $$
Simplified:
$$ 2x - 4y + 10z = 72 $$
Partial derivatives and gradient vectors are used to determine tangent planes.
No. Tangent planes exist in three-dimensional space, approximating surfaces near a point.
A tangent line touches a curve, while a tangent plane touches a surface and contains infinitely many tangent lines.
The normal line is perpendicular to the tangent plane and passes through the point of tangency.
Finding tangent planes manually requires multiple differentiation steps. The Tangent Plane Calculator automates this process, delivering fast and precise results for complex multivariable functions while maintaining mathematical accuracy.
Wikipedia – Tangent Planes
OpenStax – Tangent Planes and Linear Approximations
LibreTexts – Tangent Plane to a Surface
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