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Magazine Chapter 9: Problem 1
Find an equation of the tangent plane to the given surface at the specified point. \(z=3 y^{2}-2 x^{2}+x, \quad(2,-1,-3)\)
Short Answer
Expert verified
The equation of the tangent plane is \( z = -7x - 6y + 5 \).
Step by step solution
01
Confirm the Surface Equation
The given surface is described by the equation \( z = 3y^2 - 2x^2 + x \). The point of interest is \( (2, -1, -3) \), which lies on the surface. First, substitute \( x = 2 \), \( y = -1 \) into the equation to confirm that \( z = -3 \) is indeed correct. \[ z = 3(-1)^2 - 2(2)^2 + 2 = 3(1) - 8 + 2 = -3. \] The point is correct as \( (2, -1, -3) \) satisfies the equation.
02
Find Partial Derivatives
To find the tangent plane, calculate the partial derivatives of the function with respect to \( x \) and \( y \). These represent the slopes of the tangent plane in the \( x \) and \( y \) directions. The partial derivative of \( z \) with respect to \( x \) is\[ \frac{\partial z}{\partial x} = -4x + 1. \]The partial derivative of \( z \) with respect to \( y \) is\[ \frac{\partial z}{\partial y} = 6y. \]
03
Evaluate Partial Derivatives at the Given Point
Evaluate the partial derivatives at the point \((2, -1, -3)\). For \( \frac{\partial z}{\partial x} \) at \((2, -1)\):\[ \frac{\partial z}{\partial x} = -4(2) + 1 = -8 + 1 = -7. \] For \( \frac{\partial z}{\partial y} \) at \((2, -1)\):\[ \frac{\partial z}{\partial y} = 6(-1) = -6. \]
04
Write the Equation of the Tangent Plane
The tangent plane to the surface at the point \((x_0, y_0, z_0) = (2, -1, -3)\) is given by:\[ z - z_0 = \frac{\partial z}{\partial x}(x - x_0) + \frac{\partial z}{\partial y}(y - y_0). \]Substituting the values,\[ z + 3 = -7(x - 2) - 6(y + 1). \]Simplifying this equation:\[ z + 3 = -7x + 14 - 6y - 6. \]This further simplifies to:\[ z = -7x - 6y + 5. \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
To understand the concept behind a tangent plane, it's crucial to grasp partial derivatives. In multivariable calculus, a partial derivative represents the rate of change of a function with respect to one variable while keeping others constant.
For a function such as the surface equation given by:
\( z = 3y^2 - 2x^2 + x \),
we determine how changes in \( x \) and \( y \) individually affect \( z \).
For a function such as the surface equation given by:
\( z = 3y^2 - 2x^2 + x \),
we determine how changes in \( x \) and \( y \) individually affect \( z \).
- The partial derivative with respect to \( x \) is \( \frac{\partial z}{\partial x} = -4x + 1 \).
- The partial derivative with respect to \( y \) is \( \frac{\partial z}{\partial y} = 6y \).
Surface Equation
A surface equation in three dimensions defines a set of points that make up a surface. In this case, the surface is described by the equation \( z = 3y^2 - 2x^2 + x \).
Each pair \((x, y)\) corresponds to a specific height \( z \), which can be seen as the height above the \((x, y)\)-plane.
Each pair \((x, y)\) corresponds to a specific height \( z \), which can be seen as the height above the \((x, y)\)-plane.
- The value of \( z \) changes depending on \( x \) and \( y \). For instance, when inserting \( x = 2 \), \( y = -1 \) into the equation, we find \( z = -3 \).
- This confirms that \((2, -1, -3)\) is a point on our given surface.
Slope of Tangent Plane
The slope of a tangent plane is determined by evaluating the partial derivatives at a given point. Essentially, the slopes tell us how the surface tilts at the specific point, creating a tangent plane that just touches the surface without cutting through it.
- At point \((2, -1)\), the slope with respect to \( x \) is \( -7 \) from \( \frac{\partial z}{\partial x} = -4x + 1 \).
- Similarly, the slope with respect to \( y \) is \( -6 \) from \( \frac{\partial z}{\partial y} = 6y \).
Equation of Tangent Plane
Once the slopes at a given point are obtained via partial derivatives, we can use them to form the equation of the tangent plane.
This equation is constructed in the fashion of a linear equation, representing the relationship between differences in \( x \), \( y \), and \( z \) around the point of tangency.
The standard form is:
\[ z - z_0 = \frac{\partial z}{\partial x}(x - x_0) + \frac{\partial z}{\partial y}(y - y_0) \]
This equation is constructed in the fashion of a linear equation, representing the relationship between differences in \( x \), \( y \), and \( z \) around the point of tangency.
The standard form is:
\[ z - z_0 = \frac{\partial z}{\partial x}(x - x_0) + \frac{\partial z}{\partial y}(y - y_0) \]
- Substituting the values from our exercise, it becomes:
\[ z + 3 = -7(x - 2) - 6(y + 1) \]. - When simplified, this results in the equation:
\[ z = -7x - 6y + 5 \].
