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Chapter 13: Problem 32

Find an equation of the plane tangent to the given surface \(z=f(x, y)\) at the indicated point \(P\). $$ z=\sqrt{25-x^{2}-y^{2}}: P=(4,-3,0) $$

Short Answer

Expert verified
The tangent plane at point \(P\) is \(z = -4x + 3y + 25\).

Step by step solution

01

Understand the Surface Equation

The surface is described by the equation \(z = \sqrt{25 - x^2 - y^2}\), which represents a hemisphere with a radius of 5 centered at the origin on the xy-plane.
02

Calculate Partial Derivatives

Find the partial derivatives \(f_x\) and \(f_y\) of the surface equation. For \(f_x = \frac{d}{dx}(\sqrt{25 - x^2 - y^2})\), use the chain rule to obtain \(f_x = \frac{-x}{\sqrt{25 - x^2 - y^2}}\). Similarly, for \(f_y = \frac{d}{dy}(\sqrt{25 - x^2 - y^2})\), use the chain rule to obtain \(f_y = \frac{-y}{\sqrt{25 - x^2 - y^2}}\).
03

Evaluate Partial Derivatives at Point P

Substitute \(x = 4\), \(y = -3\) into the partial derivatives. Thus, \(f_x(4, -3) = \frac{-4}{\sqrt{25 - 4^2 - (-3)^2}} = -4\), and \(f_y(4, -3) = \frac{-(-3)}{\sqrt{25 - 4^2 - (-3)^2}} = 3\).
04

Write Equation of Tangent Plane

The general formula for a tangent plane to a surface \(z = f(x, y)\) at a point \(P(x_0, y_0, z_0)\) is: \(z - z_0 = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)\). Substitute \(x_0 = 4\), \(y_0 = -3\), \(z_0 = 0\), \(f_x = -4\), and \(f_y = 3\) into the formula to get the equation: \(z = -4(x - 4) + 3(y + 3)\).
05

Simplify Tangent Plane Equation

Simplify the equation from Step 4: \(z = -4x + 16 + 3y + 9\), or \(z = -4x + 3y + 25\). This is the equation of the tangent plane at point \(P\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Derivatives
In calculus, partial derivatives are used to measure how a function changes as its input variables change, but with respect to one variable at a time. For the function given by the surface equation \( z = \sqrt{25-x^{2}-y^{2}} \), partial derivatives \( f_x \) and \( f_y \) represent the rate of change of the function with respect to \( x \) and \( y \), respectively.
  • To find \( f_x \), differentiate \( \sqrt{25-x^2-y^2} \) with respect to \( x \).
  • The result is \( \frac{-x}{\sqrt{25-x^2-y^2}} \), telling us how \( z \) changes as \( x \) changes.
  • Similarly, \( f_y \) is obtained by differentiating with respect to \( y \).
  • This results in \( \frac{-y}{\sqrt{25-x^2-y^2}} \), describing how \( z \) changes with \( y \).
These derivatives are crucial for finding tangent planes, which provide a linear approximation of the surface near a given point.
Chain Rule
The chain rule in calculus is a method for finding the derivative of a composite function. When dealing with functions involving square roots, like in the surface equation \( z = \sqrt{25-x^2-y^2} \), partial derivatives require the chain rule.
  • The chain rule helps us differentiate nested functions by breaking them into simpler pieces.
  • In our problem, we use \( \frac{d}{du}(\sqrt{u}) = \frac{1}{2\sqrt{u}} \) to find how changes in \( x \) and \( y \) affect \( z \).
  • For \( f_x \), substitute \( u = 25-x^2-y^2 \) into the derivative formula.
  • This results in parts that are easier to manage: \( \frac{-x}{\sqrt{u}} \).
By applying the chain rule, we derive formulas that show how slight variations in inputs lead to changes in the output鈥攌ey to defining tangent planes accurately.
Hemisphere Surface
A hemisphere is a half-sphere, and it can be described using mathematical equations. The specific equation provided, \( z = \sqrt{25-x^2-y^2} \), represents a hemisphere with a radius of 5. This surface is centered at the origin on the \( xy \)-plane.
  • This formula is derived from the sphere equation \( x^2 + y^2 + z^2 = 25 \).
  • By setting \( z \) as a function of \( x \) and \( y \), we look only at the top half of the sphere, forming a hemisphere.
  • Visualizing the surface helps understand how tangent planes interact with it.
  • The tangent plane touches the hemisphere at exactly one point, giving it a linear approximation there.
Understanding the spatial representation of the hemisphere aids in comprehending how calculus tools, like partial derivatives, apply to three-dimensional surfaces.
Calculus Problem Solving
Calculus is a powerful mathematical tool for solving complex problems involving change and motion. Solving for tangent planes involves several steps, rooted in fundamental calculus concepts like derivatives.
  • Identify the function describing the surface or shape, such as \( z = \sqrt{25-x^2-y^2} \).
  • Compute partial derivatives to capture how the function changes with each variable.
  • Evaluate these derivatives at a given point to understand local behavior.
  • Construct the tangent plane equation using these derivatives and the point's coordinates.
Here, understanding each step in finding the tangent plane helps manage larger calculus problems. Mastery over such methods facilitates tackling diverse mathematical scenarios, making calculus indispensable in analyzing real-world problems.

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