Problem 45 - Calculus with Concepts in Calculus

Question

Find an equation of the plane tangent to the given surface at the given point. $$ z=\ln \sqrt{x^{2}+1} ;(0,2,0) $$

Step-by-Step Solution

Answer

The equation of the tangent plane is \( z = 0 \).

Step 1: Identify the given surface and the point of tangency

The problem provides the surface equation as \( z = \ln \sqrt{x^2 + 1} \). We need to find the tangent plane at the given point \( (0, 2, 0) \). Here, \( x = 0 \), \( y = 2 \), and \( z = 0 \) at this point.

Step 2: Calculate the partial derivatives of the surface

To find the equation of the tangent plane, we need the gradient vector of the surface. Compute the partial derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \). Start with \( \frac{\partial z}{\partial x} = \frac{d}{dx} \left( \ln \sqrt{x^2 + 1} \right) = \frac{x}{x^2 + 1} \). Since the surface does not explicitly depend on \( y \), \( \frac{\partial z}{\partial y} = 0 \).

Step 3: Evaluate the partial derivatives at the point of tangency

Substitute \( x = 0 \) and \( y = 2 \) into the partial derivatives. Since \( z = \ln \sqrt{x^2 + 1} \) only involves \( x \), \( \frac{\partial z}{\partial y} = 0 \) remains valid. Thus, \( \frac{\partial z}{\partial x} = \frac{0}{0^2 + 1} = 0 \).

Step 4: Use the point and gradient to find the tangent plane equation

The general formula for the equation of a tangent plane is: \( z = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b) \), where \( f_x \) and \( f_y \) are the partial derivatives at the point \( (a, b) \). Here, \( a = 0 \), \( b = 2 \), and \( f(a, b) = 0 \) because \( z = \ln \sqrt{0^2 + 1} = 0 \). So, the plane becomes: \( z = 0 + 0 \cdot (x - 0) + 0 \cdot (y - 2) \), which simplifies to \( z = 0 \).

Step 5: Conclusion

The equation of the plane tangent to the surface \( z = \ln \sqrt{x^2 + 1} \) at the point \( (0, 2, 0) \) is \( z = 0 \). This indicates that the tangent plane is the xy-plane.

Key Concepts

Partial Derivatives

Partial derivatives are an essential concept in calculus, especially when dealing with multivariable functions. They help us understand how a function changes as each variable is varied separately. Think of a surface described by a function like \( z = f(x, y) \). Partial derivatives tell us the rate at which \( z \) changes when only \( x \) changes while \( y \) is kept constant, and vice versa.

• To compute the partial derivative with respect to \( x \), treat \( y \) as a constant. Use the standard derivative rules for the resulting one-variable function.
• Similarly, to find the partial derivative with respect to \( y \), treat \( x \) as constant and differentiate accordingly.

Partial derivatives are crucial when determining the slope of the tangent plane to a surface at a particular point. By solving for the partial derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \), you provide information about how the surface tilts along the \( x \) and \( y \) axes, respectively.

Gradient Vector

The gradient vector encapsulates the partial derivatives of a multivariable function and provides a very intuitive geometrical interpretation. In the context of surfaces, it points in the direction of the steepest ascent.

• The gradient vector \( abla f \) for a function \( f(x, y) \) is \( \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).
• This vector not only indicates the direction but also the rate of maximum increase of the function.

When working with tangent planes, knowing the gradient at a specific point is key. It helps you establish the orientation of the tangent plane since the plane is perpendicular to this vector. This relationship is pivotal in calculating the tangent plane equation by combining the information from individual partial derivatives.

Calculus Problems

Calculus problems often involve finding solutions that require an understanding of derivatives and continuity. When grappling with these problems, you frequently encounter tasks like determining tangent planes or curves.

• These problems require recognizing the underlying function and its variables to compute derivatives accurately.
• Another essential step is substituting these derivatives into logical formulas that describe geometric objects, such as planes or tangents.

In the case of tangent planes, once you compute the derivatives (or the gradient), the goal is to use these values to write an equation for the plane. Typically, this involves using a formula for the tangent plane that incorporates these derivatives evaluated at a given point. This approach is central to solving many calculus problems related to surfaces and curves in multivariable settings.