Chapter 13: Problem 30
Use the normal gradient vector to write an equation of the line (or plane) tangent to the given curve (or surface) at the given point \(P\). $$ 2 x^{2}+3 y^{2}=35 ; \quad P(2,3) $$
Short Answer
Expert verified
The equation of the tangent line is \(8x + 18y = 70\).
Step by step solution
01
Determine the Gradient Vector
Find the gradient vector \( abla f \) of the function \( f(x, y) = 2x^2 + 3y^2 - 35 \). The gradient vector is given by \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). Calculate \( \frac{\partial f}{\partial x} = 4x \) and \( \frac{\partial f}{\partial y} = 6y \). So, \( abla f = (4x, 6y) \).
02
Evaluate the Gradient at Point P
Substitute the coordinates of point \( P(2,3) \) into the gradient vector. \( abla f(P) = (4(2), 6(3)) = (8, 18) \). This gives the normal vector at point P.
03
Write the Equation of the Tangent Line
For a curve defined implicitly, the tangent line at point \((x_0, y_0)\) is given by \((x - x_0)\frac{\partial f}{\partial x} + (y - y_0)\frac{\partial f}{\partial y} = 0\). Plugging our normal vector and point P into this gives \((x - 2)(8) + (y - 3)(18) = 0\).
04
Simplify the Equation
Expanding, we get \(8x - 16 + 18y - 54 = 0\), which simplifies to \(8x + 18y = 70\). This is the equation of the tangent line.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Tangent Plane Equation
When dealing with surfaces in calculus, a tangent plane is the plane that just "touches" a surface at a given point. It is essentially the best linear approximation of the surface around the point. To find the equation of the tangent plane, you typically need a function of the form \( f(x, y, z) = 0 \), and a point \( P(x_0, y_0, z_0) \) on the surface. The equation can be derived using the gradient vector. This vector is normal (perpendicular) to the tangent plane. If you know the gradient at a certain point, you can form the tangent plane using the formula:
\[(x - x_0)\frac{\partial f}{\partial x} + (y - y_0)\frac{\partial f}{\partial y} + (z - z_0)\frac{\partial f}{\partial z} = 0\]
\[(x - x_0)\frac{\partial f}{\partial x} + (y - y_0)\frac{\partial f}{\partial y} + (z - z_0)\frac{\partial f}{\partial z} = 0\]
- If the function is in two variables, the formula simplifies by excluding the terms involving \( z \).
- A tangent plane is a useful tool for approximating function values without the need for curves and slopes.
Implicit Differentiation
Implicit differentiation is a method to find derivatives of functions that are not isolated or explicitly solved for one variable. This technique is necessary for equations where \( y \) cannot easily be separated from \( x \). For example, when working with \( 2x^2 + 3y^2 = 35 \), you use implicit differentiation to find derivatives without solving for \( y \).
Here’s how it works:
Here’s how it works:
- Differentiate both sides of the equation with respect to \( x \). Remember to treat \( y \) as a function of \( x \), applying the chain rule when differentiating \( y \) terms.
- This results in derivatives of each term, where \( \frac{dy}{dx} \) appears for \( y \) derivatives.
Partial Derivatives
Partial derivatives represent the rate of change of a multi-variable function when all but one variable are held constant. In a function \( f(x, y) \), the partial derivative with respect to \( x \) is denoted as \( \frac{\partial f}{\partial x} \), and similarly for \( y \).
If you consider a function \( f(x, y) = 2x^2 + 3y^2 - 35 \), here’s how to find its partial derivatives:
If you consider a function \( f(x, y) = 2x^2 + 3y^2 - 35 \), here’s how to find its partial derivatives:
- To compute \( \frac{\partial f}{\partial x} \), differentiate the function with respect to \( x \) while treating \( y \) as a constant. This gives \( 4x \).
- For \( \frac{\partial f}{\partial y} \), differentiate the function with respect to \( y \), treating \( x \) as a constant, resulting in \( 6y \).
Normal Vector
In geometry, a normal vector is a vector that is perpendicular to a given surface at a specific point. The gradient vector \( abla f \) of a function \( f(x, y) \) is a normal vector to a level curve or surface.
To illustrate, consider the level curve defined by \( 2x^2 + 3y^2 = 35 \). The gradient \( abla f(x, y) = (4x, 6y) \) serves as the normal vector to this curve. At point \( P(2, 3) \), the normal vector is \( abla f(P) = (8, 18) \).
To illustrate, consider the level curve defined by \( 2x^2 + 3y^2 = 35 \). The gradient \( abla f(x, y) = (4x, 6y) \) serves as the normal vector to this curve. At point \( P(2, 3) \), the normal vector is \( abla f(P) = (8, 18) \).
- Normal vectors are essential in forming the equation of the tangent plane.
- They provide a simple way to understand the orientation of a surface at a particular point.
- They are extensively used in visual computations, such as calculating illumination in computer graphics.