Problem 5
Question
A function \(z=f(x, y),\) a vector \(\vec{v}\) and a point \(P\) are given. Give the parametric equations of the following directional tangent lines to \(f\) at \(P\) : (a) \(\ell_{x}(t)\) (b) \(\ell_{y}(t)\) (c) \(\ell_{\vec{u}}(t),\) where \(\vec{u}\) is the unit vector in the direction of \(\vec{v}\). $$ f(x, y)=2 x^{2} y-4 x y^{2}, \vec{v}=\langle 1,3\rangle, P=(2,3) $$
Step-by-Step Solution
Verified Answer
\(\ell_x(t) = (2+t, 3, 18-12t), \ell_y(t) = (2, 3+t, 18-40t), \ell_{\vec{u}}(t) = (2+\frac{t}{\sqrt{10}}, 3+\frac{3t}{\sqrt{10}}, 18-\frac{132t}{\sqrt{10}})\).
1Step 1: Calculate Partial Derivatives
Find the partial derivatives of the function \(z=f(x, y)\) with respect to \(x\) and \(y\). For \(f(x, y) = 2x^2y - 4xy^2\), the partial derivative with respect to \(x\) is \(\frac{\partial f}{\partial x} = 4xy - 4y^2\). The partial derivative with respect to \(y\) is \(\frac{\partial f}{\partial y} = 2x^2 - 8xy\).
2Step 2: Evaluate Partial Derivatives at Point P
Substitute \(P = (2, 3)\) into the partial derivatives. For \(\frac{\partial f}{\partial x}\), substitute \(x = 2, y = 3\), yielding \(4(2)(3) - 4(3)^2 = 24 - 36 = -12\). For \(\frac{\partial f}{\partial y}\), substitute \(x = 2, y = 3\), yielding \(2(2)^2 - 8(2)(3) = 8 - 48 = -40\).
3Step 3: Parametric Equation for \(\ell_x(t)\)
The directional tangent line in the \(x\)-direction at \(P\) is given by \(\ell_x(t) = (x_0 + t, y_0, f(x_0, y_0) + t \frac{\partial f}{\partial x}(x_0, y_0))\). Substituting \(x_0 = 2, y_0 = 3\) and \(\frac{\partial f}{\partial x} = -12\), the parametric equation is \(\ell_x(t) = (2+t, 3, 18 - 12t)\) as \(f(2, 3) = 18\).
4Step 4: Parametric Equation for \(\ell_y(t)\)
The directional tangent line in the \(y\)-direction at \(P\) is given by \(\ell_y(t) = (x_0, y_0 + t, f(x_0, y_0) + t \frac{\partial f}{\partial y}(x_0, y_0))\). Using \(f(x_0, y_0) = 18\), \(\frac{\partial f}{\partial y} = -40\), \(\ell_y(t) = (2, 3+t, 18 - 40t)\).
5Step 5: Find the Unit Vector \(\vec{u}\)
Calculate the unit vector \(\vec{u}\) in the direction of \(\vec{v} = \langle 1, 3 \rangle\). The magnitude of \(\vec{v}\) is \(||\vec{v}|| = \sqrt{1^2 + 3^2} = \sqrt{10}\). Thus, \(\vec{u} = \langle \frac{1}{\sqrt{10}}, \frac{3}{\sqrt{10}} \rangle\).
6Step 6: Parametric Equation for \(\ell_\vec{u}(t)\)
The directional tangent line in the direction of \(\vec{u}\) is obtained by \(\ell_{\vec{u}}(t) = (x_0 + t \cdot \frac{1}{\sqrt{10}}, y_0 + t \cdot \frac{3}{\sqrt{10}}, f(x_0, y_0) + t(\frac{\partial f}{\partial x}\frac{1}{\sqrt{10}} + \frac{\partial f}{\partial y}\frac{3}{\sqrt{10}}))\). Substituting values, \(\ell_{\vec{u}}(t) = (2 + \frac{t}{\sqrt{10}}, 3 + \frac{3t}{\sqrt{10}}, 18 + t(-\frac{12}{\sqrt{10}} - \frac{120}{\sqrt{10}})) = (2 + \frac{t}{\sqrt{10}}, 3 + \frac{3t}{\sqrt{10}}, 18 - \frac{132t}{\sqrt{10}})\).
Key Concepts
Partial DerivativesTangent LinesParametric EquationsUnit Vector
Partial Derivatives
Partial derivatives are like slices of a multi-dimensional function that show how the function changes as each variable changes, while all other variables are kept constant. For a function of two variables, like \(z = f(x, y)\), there are two primary partial derivatives: one with respect to \(x\) and one with respect to \(y\). These derivatives help us understand the "slope" or rate of change of the function in the direction of the respective variable.
- **Partial derivative with respect to \(x\)**: This is found by differentiating \(f(x, y)\) with \(y\) held constant. It tells us how \(f\) changes as only \(x\) changes. In our example, \(\frac{\partial f}{\partial x} = 4xy - 4y^2\).
- **Partial derivative with respect to \(y\)**: Similarly, this is obtained by differentiating \(f(x, y)\) while keeping \(x\) constant. It indicates how \(f\) changes with changes in \(y\). We calculated \(\frac{\partial f}{\partial y} = 2x^2 - 8xy\).
Evaluating these partial derivatives at a specific point, like \(P = (2, 3)\), gives the actual slopes of the function at that point in the \(x\) and \(y\) directions respectively. This evaluation is key in finding directional derivatives and constructing tangent lines.
- **Partial derivative with respect to \(x\)**: This is found by differentiating \(f(x, y)\) with \(y\) held constant. It tells us how \(f\) changes as only \(x\) changes. In our example, \(\frac{\partial f}{\partial x} = 4xy - 4y^2\).
- **Partial derivative with respect to \(y\)**: Similarly, this is obtained by differentiating \(f(x, y)\) while keeping \(x\) constant. It indicates how \(f\) changes with changes in \(y\). We calculated \(\frac{\partial f}{\partial y} = 2x^2 - 8xy\).
Evaluating these partial derivatives at a specific point, like \(P = (2, 3)\), gives the actual slopes of the function at that point in the \(x\) and \(y\) directions respectively. This evaluation is key in finding directional derivatives and constructing tangent lines.
Tangent Lines
Tangent lines provide a linear approximation of surfaces at a specific point, almost like giving a surface a flat path to follow locally. When dealing with a function \(z = f(x, y)\), tangent lines in the directions of \(x\) and \(y\) offer insights into the surface's local linear behavior.
- **\(x\)-direction tangent line (\(\ell_x(t)\))**: Formed by moving in the \(x\)-direction from point \((x_0, y_0)\), keeping \(y\) fixed. It's represented as \(\ell_x(t) = (x_0 + t, y_0, f(x_0, y_0) + t \frac{\partial f}{\partial x}(x_0, y_0))\). In our exercise, this becomes \((2+t, 3, 18 - 12t)\).
- **\(y\)-direction tangent line (\(\ell_y(t)\))**: Here, \(x\) is held steady and \(y\) changes. This line is \(\ell_y(t) = (x_0, y_0 + t, f(x_0, y_0) + t \frac{\partial f}{\partial y}(x_0, y_0))\), which results in \((2, 3+t, 18 - 40t)\) for this function.
These tangible linear paths provide a way to visualize how a complex surface behaves nearby a specific point, telling us how the value of the function would grow or shrink moving in the cardinal directions.
- **\(x\)-direction tangent line (\(\ell_x(t)\))**: Formed by moving in the \(x\)-direction from point \((x_0, y_0)\), keeping \(y\) fixed. It's represented as \(\ell_x(t) = (x_0 + t, y_0, f(x_0, y_0) + t \frac{\partial f}{\partial x}(x_0, y_0))\). In our exercise, this becomes \((2+t, 3, 18 - 12t)\).
- **\(y\)-direction tangent line (\(\ell_y(t)\))**: Here, \(x\) is held steady and \(y\) changes. This line is \(\ell_y(t) = (x_0, y_0 + t, f(x_0, y_0) + t \frac{\partial f}{\partial y}(x_0, y_0))\), which results in \((2, 3+t, 18 - 40t)\) for this function.
These tangible linear paths provide a way to visualize how a complex surface behaves nearby a specific point, telling us how the value of the function would grow or shrink moving in the cardinal directions.
Parametric Equations
Parametric equations express a set of quantities as explicit functions of a parameter, often time \(t\). In the context of directional derivatives and tangent lines, parametric equations describe the path a tangent takes, giving a simple way to trace the curve of the line in the \(xyz\)-space.
- **Structure of parametric equations**: Each coordinate (\(x\), \(y\), \(z\)) is made a function of \(t\). For example, in our \(x\)-direction tangent line, \(\ell_x(t) = (x_0 + t, y_0, f(x_0, y_0) + t\frac{\partial f}{\partial x})\).
- **Benefits**: They simplify understanding and calculating paths, as all movements and slopes are expressed relative to \(t\). This makes it intuitive to plug in different \(t\) values and visualize how the function extends or contracts in space.
By substituting different values of \(t\), students can see how these tangent lines evolve, providing a concrete grasp on abstract mathematical concepts like directionality and changes in multidimensional contexts.
- **Structure of parametric equations**: Each coordinate (\(x\), \(y\), \(z\)) is made a function of \(t\). For example, in our \(x\)-direction tangent line, \(\ell_x(t) = (x_0 + t, y_0, f(x_0, y_0) + t\frac{\partial f}{\partial x})\).
- **Benefits**: They simplify understanding and calculating paths, as all movements and slopes are expressed relative to \(t\). This makes it intuitive to plug in different \(t\) values and visualize how the function extends or contracts in space.
By substituting different values of \(t\), students can see how these tangent lines evolve, providing a concrete grasp on abstract mathematical concepts like directionality and changes in multidimensional contexts.
Unit Vector
A unit vector is a vector of length 1 that maintains direction. In directional derivatives, unit vectors are crucial because they standardize the direction without altering the magnitude of change.
- **Constructing a unit vector**: From any vector \(\vec{v} = \langle a, b \rangle\), its unit vector \(\vec{u}\) is found by dividing each component by the vector's magnitude. The magnitude is given by \(||\vec{v}|| = \sqrt{a^2 + b^2}\).
- **Example**: Given \(\vec{v} = \langle 1, 3 \rangle\), its magnitude is \(\sqrt{1^2 + 3^2} = \sqrt{10}\), so \(\vec{u} = \langle \frac{1}{\sqrt{10}}, \frac{3}{\sqrt{10}} \rangle\).
Unit vectors allow us to describe the direction concisely when finding a tangent line or a derivative, ensuring the calculations focus purely on direction while keeping the scale neutral. They enable straightforward calculations of dot products, such as in the formulation of directional derivatives, and help develop a consistent approach in multi-dimensional analysis.
- **Constructing a unit vector**: From any vector \(\vec{v} = \langle a, b \rangle\), its unit vector \(\vec{u}\) is found by dividing each component by the vector's magnitude. The magnitude is given by \(||\vec{v}|| = \sqrt{a^2 + b^2}\).
- **Example**: Given \(\vec{v} = \langle 1, 3 \rangle\), its magnitude is \(\sqrt{1^2 + 3^2} = \sqrt{10}\), so \(\vec{u} = \langle \frac{1}{\sqrt{10}}, \frac{3}{\sqrt{10}} \rangle\).
Unit vectors allow us to describe the direction concisely when finding a tangent line or a derivative, ensuring the calculations focus purely on direction while keeping the scale neutral. They enable straightforward calculations of dot products, such as in the formulation of directional derivatives, and help develop a consistent approach in multi-dimensional analysis.
Other exercises in this chapter
Problem 4
T/F: Along a level curve, the output of a function does not change.
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Find the critical points of the given function. Use the Second Derivative Test to determine if each critical point corresponds to a relative maximum, minimum, o
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The gradient points in the direction of __________ increase.
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In Exercises \(5-8,\) find the total differential \(d z\). $$ z=x \sin y+x^{2} $$
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