Problem 34 - Calculus with Concepts in Calculus

Question

Find an equation of the plane tangent to the graph of the given function at the indicated point(s). $$ f(x, y)=e^{x+y^{2}} ;\left(-1,0, e^{-1}\right) \text { and }(0,1, e) $$

Step-by-Step Solution

Answer

Tangent plane equations: \(z = e^{-1}x + e^{-1}\) at \((-1,0,e^{-1})\) and \(z = ex + 2ey - e\) at \((0,1,e)\).

Step 1: Understand the Problem

We need to find the equation of the tangent plane for a function of two variables, \(f(x, y) = e^{x + y^2}\), at the given points, \((-1,0, e^{-1})\) and \((0,1, e)\). To do this, we will use the point-slope equation of a plane for functions of two variables.

Step 2: Calculate Partial Derivatives

Calculate the partial derivatives of \(f(x, y)\). \[ \frac{\partial f}{\partial x} = \frac{d}{dx} e^{x + y^2} = e^{x + y^2} \]\[ \frac{\partial f}{\partial y} = \frac{d}{dy} e^{x + y^2} = 2ye^{x + y^2} \]

Step 3: Apply Partial Derivatives at the First Point

For the point \((-1,0,e^{-1})\), calculate:\[ \frac{\partial f}{\partial x}(-1,0) = e^{-1 + 0^2} = e^{-1} \]\[ \frac{\partial f}{\partial y}(-1,0) = 2(0)e^{-1 + 0^2} = 0 \]

Step 4: Find the Tangent Plane Equation at First Point

Use the point \((-1, 0, e^{-1})\) and the derivatives to find the equation: The equation of the tangent plane is given by \[ z - f(x_0, y_0) = \frac{\partial f}{\partial x}(x_0, y_0)(x - x_0) + \frac{\partial f}{\partial y}(x_0, y_0)(y - y_0) \]Substituting \(x_0 = -1\), \(y_0 = 0\), and \(z_0 = e^{-1}\):\[ z - e^{-1} = e^{-1}(x + 1) + 0(y - 0) \]Simplifying, we get: \[ z = e^{-1}(x + 1) + e^{-1} = e^{-1}x + e^{-1} \]

Step 5: Apply Partial Derivatives at the Second Point

For the point \((0,1,e)\), calculate:\[ \frac{\partial f}{\partial x}(0,1) = e^{0 + 1^2} = e \]\[ \frac{\partial f}{\partial y}(0,1) = 2(1)e^{0 + 1^2} = 2e \]

Step 6: Find the Tangent Plane Equation at Second Point

Use the point \((0, 1, e)\) and the derivatives to find the equation:Substituting \(x_0 = 0\), \(y_0 = 1\), and \(z_0 = e\):\[ z - e = e(x - 0) + 2e(y - 1) \]Simplifying, we get:\[ z = ex + 2ey + e - 2e = ex + 2ey - e \]

Step 7: Conclusion

We have two tangent planes:1. At point \((-1, 0, e^{-1})\), the equation of the tangent plane is \(z = e^{-1}x + e^{-1}\).2. At point \((0, 1, e)\), the equation of the tangent plane is \(z = ex + 2ey - e\).

Key Concepts

Partial Derivatives

In multivariable calculus, understanding partial derivatives is crucial for exploring how functions change. A function like \(f(x, y) = e^{x + y^2}\) deals with more than one variable. To find out how changes in one specific variable affect the function, while keeping others constant, we use partial derivatives.

With partial derivatives:

• \(\frac{\partial f}{\partial x}\) is the partial derivative with respect to \(x\), treating \(y\) as a constant.
• \(\frac{\partial f}{\partial y}\) is the partial derivative with respect to \(y\), now treating \(x\) as a constant.

When we calculate \(\frac{\partial f}{\partial x}\) for our function, we treat \(y^2\) as a constant part of the expression. This results in \( e^{x+y^2} \). Similarly, solving for \(\frac{\partial f}{\partial y}\) gives \(2ye^{x+y^2}\), reflecting how the \(y\) variable affects the function differently than \(x\) does.

Both derivatives are essential for forming tangent planes because they effectively describe the slope of the surface in relation to each axis. Understanding them is key to grasping more complex multivariable calculus problems.

Multivariable Calculus

Multivariable calculus extends the principles of single-variable calculus to functions of several variables. It provides powerful tools for understanding how changes in multiple dimensions affect a function.

Among these tools, the concept of a tangent plane is particularly important. A tangent plane approximates the surface of a multivariable function at a specific point.

• It is analogous to a tangent line in single-variable calculus, but extended to two or more dimensions.
• Tangent planes give us a way to estimate the values of a function near a point using linear expressions.

For our function, \(f(x, y) = e^{x + y^2}\), calculating the tangent planes at specified points involves finding partial derivatives first. Once calculated, these derivatives help in composing a linear equation which serves as the tangent plane.

By calculating tangent planes, we understand both the direction and rate at which changes occur on the surface, at those given points. This makes multivariable calculus an indispensable tool in fields that require dimensional analysis of complex systems.

Point-Slope Equation

The point-slope form is fundamental in finding equations of tangent planes in multivariable calculus. While familiar in two dimensions, it adapts well to three-dimensional spaces. In the context of tangent planes:

• The general format is \(z - z_0 = a(x - x_0) + b(y - y_0)\).
• \(z_0\) is the value of the function at \((x_0, y_0)\), while \(a\) and \(b\) are the partial derivatives at that point.

This equation forms a plane that touches the surface of the function exactly at point \((x_0, y_0, z_0)\).

For instance, at point \((-1, 0, e^{-1})\) with the function \(f(x, y) = e^{x + y^2}\), the partial derivatives \(a = e^{-1}\) and \(b = 0\) help shape the tangent plane equation: \(z = e^{-1}x + e^{-1}\). Similarly, at point \((0, 1, e)\), using \(a = e\) and \(b = 2e\), we derive the plane \(z = ex + 2ey - e\).

This application of the point-slope equation is key in dynamically understanding multiple surfaces and their behavior around specific points.