Chapter 13: Problem 34
Find an equation of the tangent plane and find symmetric equations of the normal line to the surface at the given point.\(x y z=10, \quad(1,2,5)\)
Short Answer
Expert verified
The equation of the tangent plane is \(10(x - 1) + 5(y - 2) + 2(z - 5) = 0\) and the symmetric equations of the normal line are \((x - 1) / 10 = (y - 2) / 5 = (z - 5) / 2\)
Step by step solution
01
Calculate the gradient
Firstly, we calculate the gradient of the function which gives the normal vector. The function is given by \(f(x,y,z) = x y z - 10\). The gradient of \(f\) is given by \(\nabla f = \\) where \(f_x, f_y, f_z\) are the partial derivatives of \(f\). In our case, \(f_x = yz\), \(f_y = xz\), and \(f_z = xy\). Substituting our point \((1,2,5)\), we get \(\nabla f(1,2,5) = \(2*5,1*5,1*2) = \<10, 5, 2>\)
02
Find the equation of the tangent plane
The equation of the tangent plane is given by \(f_x(x - a) + f_y(y - b) + f_z(z - c) = 0\) where \((a,b,c)\) is our given point and \((f_x, f_y, f_z)\) is the gradient vector we calculated at the given point. Substituting our values we have: \(10(x - 1) + 5(y - 2) + 2(z - 5) = 0\)
03
Find symmetric equations of the normal line
The symmetric equations of the normal line to the surface at the given point is given by \((x - x_0) / A = (y - y_0) / B = (z - z_0) / C\), where A, B and C are the components of the normal vector to the tangent plane and \((x_0, y_0, z_0)\) is our given point. Substituting our values we get \((x - 1) / 10 = (y - 2) / 5 = (z - 5) / 2\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gradient
The gradient of a function is an essential concept in calculus that acts like a compass pointing in the direction of the greatest rate of increase of the function. For multivariable functions, the gradient is a vector that consists of all its first partial derivatives. When you have a function like the one given, \[ f(x, y, z) = x y z - 10 \] we calculate each of the partial derivatives. These are:
- \( f_x = yz \) - this is the derivative of \( f \) with respect to \( x \)
- \( f_y = xz \) - this is the derivative of \( f \) with respect to \( y \)
- \( f_z = xy \) - this is the derivative of \( f \) with respect to \( z \)
Normal Line
The normal line to a surface at a given point is a line that is perpendicular to the tangent plane of the surface at that point. When finding the normal line, the key is to use the gradient vector, \( abla f \), calculated from the previous section. This gradient directly acts as the direction vector for our normal line. To express the symmetric equations of the normal line, you take the formula: \[ \frac{x - x_0}{A} = \frac{y - y_0}{B} = \frac{z - z_0}{C} \]. Here, \( (x_0, y_0, z_0) \) is the point, \( (1, 2, 5) \), and \( A, B, C \) are the components of the gradient vector \( \langle 10, 5, 2 \rangle \). So, after substituting these into the equation, we get: \[ \frac{x - 1}{10} = \frac{y - 2}{5} = \frac{z - 5}{2} \]. This describes a line in three-dimensional space that is normal to the surface at the point (1, 2, 5). It extends infinitely in both directions, providing a clear path along the steepest ascent from the surface.
Partial Derivatives
Partial derivatives are used to measure how a function changes as only one of the input variables changes, while others are kept constant. For a function of multiple variables such as \( f(x, y, z) \), you calculate the partial derivatives with respect to each variable: \( x \), \( y \), and \( z \). This is similar to finding a derivative in single-variable calculus but applied to each variable individually.
- For \( x \), the partial derivative is \( f_x = yz \), meaning it shows how changes in \( x \) affect \( f \), with \( y \) and \( z \) constant.
- For \( y \), the partial derivative is \( f_y = xz \), indicating how changes in \( y \) affect \( f \) while holding \( x \) and \( z \) steady.
- For \( z \), it is \( f_z = xy \), revealing how changes in \( z \) impact \( f \) when \( x \) and \( y \) remain unchanged.