Problem 23
Question
An implicitly defined function of \(x, y\) and \(z\) is given along with a point \(P\) that lies on the surface. Use the gradient \(\nabla F\) to: (a) find the equation of the normal line to the surface at \(P,\) and (b) find the equation of the plane tangent to the surface at \(P\). \(x y^{2}-x z^{2}=0,\) at \(P=(2,1,-1)\)
Step-by-Step Solution
Verified Answer
Normal line: \( x = 2, y = 1 + 4t, z = -1 + 4t \); Tangent plane: \( y + z = 0 \).
1Step 1: Find the Gradient of F
The function given is implicitly defined as \( F(x, y, z) = xy^2 - xz^2 = 0 \). We need to find the gradient \( abla F \), which consists of the partial derivatives of \( F \) with respect to \( x, y, \) and \( z \). So, compute these partial derivatives:\[ \frac{\partial F}{\partial x} = y^2 - z^2 \]\[ \frac{\partial F}{\partial y} = 2xy \]\[ \frac{\partial F}{\partial z} = -2xz \]
2Step 2: Evaluate Gradient at Point P
Next, evaluate the gradient \( abla F \) at the point \( P = (2, 1, -1) \):- \( \left. \frac{\partial F}{\partial x} \right|_{(2, 1, -1)} = 1^2 - (-1)^2 = 0 \)- \( \left. \frac{\partial F}{\partial y} \right|_{(2, 1, -1)} = 2 \cdot 2 \cdot 1 = 4 \)- \( \left. \frac{\partial F}{\partial z} \right|_{(2, 1, -1)} = -2 \cdot 2 \cdot (-1) = 4 \)Thus, \( abla F(2, 1, -1) = \langle 0, 4, 4 \rangle \).
3Step 3: Equation of the Normal Line
The normal line to the surface at \(P\) is in the direction of the gradient vector evaluated at \(P\). Use the formula for the line: \[ \mathbf{r}(t) = \mathbf{r}_0 + tabla F(2, 1, -1) \]Where \( \mathbf{r}_0 = (2, 1, -1) \) and \( abla F(2, 1, -1) = \langle 0, 4, 4 \rangle \):\[ x = 2, \quad y = 1 + 4t, \quad z = -1 + 4t \]
4Step 4: Equation of the Tangent Plane
The tangent plane to the surface at \( P \) is given by the equation: \[ abla F(2, 1, -1) \cdot (x - 2, y - 1, z + 1) = 0 \]Expand and simplify the dot product:\[ 0(x - 2) + 4(y - 1) + 4(z + 1) = 0 \]This simplifies to:\[ 4y + 4z = 0 \] or \[ y + z = 0 \].
5Step 5: Conclusion
The normal line has direction \( \langle 0, 4, 4 \rangle \) and passes through \( P \). The tangent plane is described by the equation \( y + z = 0 \).
Key Concepts
Gradient VectorTangent PlaneNormal LinePartial Derivatives
Gradient Vector
Gradient vectors are essential in calculus for finding directions in multivariable functions. When dealing with functions like the one in this exercise, where the function is implicitly defined as \( F(x, y, z) = xy^2 - xz^2 = 0 \), the gradient vector \( abla F \) consists of all the partial derivatives of \( F \).
- The partial derivative with respect to \(x\) is \( \frac{\partial F}{\partial x} = y^2 - z^2 \).
- The partial derivative with respect to \(y\) is \( \frac{\partial F}{\partial y} = 2xy \).
- The partial derivative with respect to \(z\) is \(\frac{\partial F}{\partial z} = -2xz\).
Tangent Plane
The tangent plane is a linear approximation of a surface at a given point, and its equation can be derived using the gradient vector. In our exercise, the equation of the tangent plane at point \(P\) is formed by using the dot product of the gradient vector evaluated at \(P\), \(abla F(2, 1, -1) = \langle 0, 4, 4 \rangle \), with a vector \((x-2, y-1, z+1)\) pointing from \(P\) to any point \((x, y, z)\) on the plane.To derive the equation, we set up the dot product expression: \[ abla F(2, 1, -1) \cdot (x - 2, y - 1, z + 1) = 0 \]When you expand and solve the dot product, you find: - \(0(x - 2)\) does not influence the outcome - \(4(y - 1) + 4(z + 1) = 0\), simplifying to \(4y + 4z = 0\), or simply \(y + z = 0\). This equation represents the tangent plane to the surface at point \(P = (2, 1, -1)\). It captures the first-order changes around \(P\), providing a good approximation to the actual surface there.
Normal Line
The normal line is perpendicular to the tangent plane at a given point on a surface. In this task, it passes through the point \(P\) and is aligned with the gradient vector \(abla F\) you've already calculated. The direction of the normal line is given by the vector \(\langle 0, 4, 4 \rangle\), calculated at \(P = (2, 1, -1)\). To express the normal line mathematically, we use a parametric form:
- \( x = 2 + 0t = 2 \)
- \( y = 1 + 4t \)
- \( z = -1 + 4t \)
Partial Derivatives
Partial derivatives measure how a function changes as each of its input variables is varied, holding the other variables constant. They are the building blocks for understanding gradients, which are vectors of partial derivatives. For the implicit function given, \( F(x, y, z) = xy^2 - xz^2 \), we calculated the following partial derivatives:
- With respect to \(x: \frac{\partial F}{\partial x} = y^2 - z^2 \)
- With respect to \(y: \frac{\partial F}{\partial y} = 2xy \)
- With respect to \(z: \frac{\partial F}{\partial z} = -2xz \)
Other exercises in this chapter
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