Problem 8
Question
The tangent plane at the indicated point \(\left(x_{0}, y_{0}, z_{0}\right)\) exists. Find its equation. $$ f(x, y)=e^{x} \cos y ;(0,0,1) $$
Step-by-Step Solution
Verified Answer
The equation of the tangent plane is \(z = x + 1\).
1Step 1: Understand the Tangent Plane Formula
The equation of the tangent plane to the surface defined by \( z = f(x, y) \) at a point \((x_0, y_0, z_0)\) is given by: \( z - z_0 = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) \), where \(f_x\) and \(f_y\) are the partial derivatives of \(f\) with respect to \(x\) and \(y\).
2Step 2: Calculate Partial Derivatives
Find the partial derivatives of \(f(x, y) = e^x \cos y\). First, the partial derivative with respect to \(x\) is \(f_x(x, y) = e^x \cos y\). Next, the partial derivative with respect to \(y\) is \(f_y(x, y) = -e^x \sin y\).
3Step 3: Evaluate Derivatives at Given Point
Evaluate these derivatives at the point \((x_0, y_0) = (0, 0)\). Thus, \(f_x(0, 0) = e^0 \cos 0 = 1\) and \(f_y(0, 0) = -e^0 \sin 0 = 0\).
4Step 4: Write the Tangent Plane Equation
Using the tangent plane formula and the derivatives evaluated, substitute into the equation: \( z - 1 = 1 \cdot (x - 0) + 0 \cdot (y - 0)\). Simplifying gives \( z - 1 = x\) or rearranged, \( z = x + 1\).
Key Concepts
Partial DerivativesTangent Plane EquationMultivariable Calculus
Partial Derivatives
Partial derivatives are crucial in multivariable calculus because they allow us to see how a function changes as we tweak one variable at a time, keeping the others fixed.
- Think of partial derivatives as a way of "slicing" through a multidimensional surface to understand its directional change.
- For example, if you have a mountain landscape described by a function, the partial derivative with respect to one variable helps you see the incline when you walk in a straight line in one specific direction.
- In mathematical terms, for a function like \(f(x, y) = e^x \cos y\), the partial derivatives are \(f_x(x, y) = e^x \cos y\) and \(f_y(x, y) = -e^x \sin y\).
Tangent Plane Equation
The tangent plane provides a linear approximation to a surface at a given point, essentially acting like a flat surface that 'touches' the curve or surface at the specific point.
- This mathematical tool helps in making complex surfaces easier to work with, especially near the point of tangency.
- The general formula for the tangent plane to a surface \(z = f(x, y)\) at a point \((x_0, y_0, z_0)\) is: \(z - z_0 = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)\).
- Here, \(f_x\) and \(f_y\) are partial derivatives, as we calculated earlier, and these help define the slope of the plane in both the x- and y-directions.
Multivariable Calculus
Multivariable calculus expands upon single-variable calculus, allowing us to explore functions of two or more variables. This includes thinking through problems that involve three-dimensional space and beyond.
- Imagine you're working with surfaces in three dimensions, such as mountains, wtih each point on the surface represented by a combination of x, y, and the function \(z=f(x,y)\).
- Concepts like partial derivatives are used to study how these variables interact and change together.
- One of the critical applications in multivariable calculus is the use of approximations like tangent planes to simplify analysis and computations.
Other exercises in this chapter
Problem 8
Find \(\frac{d w}{d t}\) for \(w=e^{f(x, y)}\) with \(x=u(t)\) and \(y=v(t)\).
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Evaluate each function at the given point. (a) \(f_{1}(x, y)=\frac{x}{y}\) at \((3,2)\) (b) \(f_{2}(y, x)=\frac{x}{y}\) at \((3,2)\) (c) \(f_{3}(y, x)=\frac{y}{
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In Problems 1-16, find \(\partial f / \partial x\) and \(\partial f / \partial y\) for the given functions. $$ f(x, y)=\sec \left(y^{2} x-x^{3}\right) $$
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