Exam Information:
Exam 1 is scheduled for February 25th at ELLT 116 from 6:30pm to 7:30pm.
Important resources posted on Brightspace include instructions, seating chart, and a study guide.
Office Hours Schedule (02/16 - 02/25):
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Given \(f(x, y, z) = z \ln(xy)\) and unit vector \(\vec{u} = \left\langle \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right\rangle\), find the directional derivative at the point \(\left(\frac{1}{4}, 4, 1\right)\).
Solution:
First, compute the gradient of \(f\):
\[\nabla f(x, y, z) = \left\langle f_x, f_y, f_z \right\rangle\]Calculate each partial derivative:
\[f_x = \frac{z}{x}\] \[f_y = \frac{z}{y}\] \[f_z = \ln(xy)\]Evaluate the gradient at \(\left(\frac{1}{4}, 4, 1\right)\):
\[\nabla f\left(\frac{1}{4}, 4, 1\right) = \left\langle 4, \frac{1}{4}, 0 \right\rangle\]The directional derivative is:
\[D_{\vec{u}}f = \nabla f \cdot \vec{u} = \left\langle 4, \frac{1}{4}, 0 \right\rangle \cdot \left\langle \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right\rangle\] \[= \frac{1}{\sqrt{3}}\left(4 + \frac{1}{4}\right) = \frac{1}{\sqrt{3}} \cdot \frac{17}{4}\]Important: The maximum rate of increase at a point occurs in the direction of the gradient vector \(\nabla f\).
The maximum rate of increase at \(\left(\frac{1}{4}, 4, 1\right)\) is:
\[|\nabla f| = \left|\left\langle 4, \frac{1}{4}, 0 \right\rangle\right| = \sqrt{16 + \frac{1}{16}} = \sqrt{\frac{257}{16}}\]
Diagram Description: A coordinate system showing a curve \(y = f(x)\) in blue. At point \(x = a\), a tangent line in purple touches the curve, and the function value \(f(a)\) is marked. The tangent line has slope \(f'(a)\).
For a curve \(y = f(x)\) in two dimensions, we approximate the function using the tangent line through the point \((a, f(a))\) with slope \(f'(a)\).
Equation of Tangent Line:
\[y = f(a) + f'(a)(x - a)\]Linear Approximation:
The linear approximation of \(f(x)\) at \(x = a\) is:
\[L(x) = f(a) + f'(a)(x - a)\]Differential:
The differential \(dy\) approximates the change in \(y\):
\[dy \approx f'(a) \, dx\]
Approximate the value of \(\sqrt{4.01}\). The exact value is approximately 2.002498.
Solution:
Let \(f(x) = \sqrt{x}\). We use linear approximation at \(x = 4\).
First, find the derivative:
\[f'(x) = \frac{1}{2\sqrt{x}}\]At \(x = 4\):
\[f(4) = 2\] \[f'(4) = \frac{1}{2\sqrt{4}} = \frac{1}{4}\]The linear approximation is:
\[L(x) = f(4) + f'(4)(x - 4) = 2 + \frac{1}{4}(x - 4)\]Evaluate at \(x = 4.01\):
\[\sqrt{4.01} \approx L(4.01) = 2 + \frac{1}{4}(0.01) = 2.0025\]Note: The approximate value 2.0025 is very close to the exact value 2.002498, highlighted in green on the page.
Diagram Description: A three-dimensional surface (shown in blue) with multiple tangent lines shown in different colors (red, purple, green) emanating from a common point on the surface. A tangent plane containing all these tangent lines is illustrated, and a normal vector \(\vec{n}\) perpendicular to the plane is shown in pink.
For a surface \(z = f(x, y)\) in three dimensions, the tangent plane at a point contains all the tangent lines to curves passing through that point.
Normal Vector: The normal vector is orthogonal (perpendicular) to each of the tangent lines and thus orthogonal to the tangent plane.
Equation of Tangent Plane:
The equation of the tangent plane to \(z = f(x, y)\) at point \((a, b, f(a,b))\) is:
\[z = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b)\]Linear Approximation in Two Variables:
The linear approximation of \(f(x, y)\) at \((a, b)\) is:
\[L(x, y) = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b)\]Differential:
The differential \(dz\) approximates the change in \(z\):
\[dz \approx f_x \, dx + f_y \, dy\]
For a surface \(z = f(x, y)\) in 3D, the tangent plane at \((a, b, f(a,b))\) is a plane containing all the tangent vectors to curves passing through that point.
Normal Vector: A vector perpendicular to all tangent vectors on the surface at the given point.
Equation of Tangent Plane:
\[z = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b)\]Linear Approximation:
\[L(x, y) = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b)\]Differential:
\[dz = f_x \, dx + f_y \, dy\]
Find the tangent plane at \((2, 3, 12)\) and approximate \(f(2.01, 2.98)\).
Solution:
First, find the partial derivatives:
\[f_x = -2x\] \[f_y = -2y\]Evaluate at \((2, 3)\):
\[f_x(2, 3) = -4\] \[f_y(2, 3) = -6\]The equation of the tangent plane is:
\[z = f(2, 3) + f_x(2, 3)(x - 2) + f_y(2, 3)(y - 3)\] \[z = 12 - 4(x - 2) - 6(y - 3)\]The linear approximation is:
\[L(x, y) = 12 - 4(x - 2) - 6(y - 3)\]Approximate \(f(2.01, 2.98)\):
\[f(2.01, 2.98) \approx L(2.01, 2.98) = 12 - 4(2.01 - 2) - 6(2.98 - 3)\] \[= 12 - 4(0.01) - 6(-0.02)\] \[= 12 - 0.04 + 0.12 = 12.08\]Therefore, \(\Delta z = 0.08\).
Let \(z = x^3 y\). Find the approximate change in \(z\) when \((x, y)\) changes from \((1, 3)\) to \((0.98, 3.04)\).
Diagram Description: Two points are shown connected by a line: \((1, 3)\) in blue and \((0.98, 3.04)\) in green, with arrows indicating \(\Delta x = -0.02\) and \(\Delta y = 0.04\).
Solution:
Use the differential formula:
\[dz = f_x \, dx + f_y \, dy\]Calculate the changes:
\[\Delta x = 0.98 - 1 = -0.02\] \[\Delta y = 3.04 - 3 = 0.04\]Find the partial derivatives:
\[f_x = 3x^2 y \implies f_x(1, 3) = 9\] \[f_y = x^3 \implies f_y(1, 3) = 1\]Calculate \(\Delta z\):
\[\Delta z \approx 9(-0.02) + 1(0.04) = -0.18 + 0.04 = -0.14\]Since \(f(1, 3) = 3\), the linear approximation at \((0.98, 3.04)\) is:
\[f(0.98, 3.04) \approx f(1, 3) + \Delta z = 3 - 0.14 = 2.86\]
A plane with equation \(ax + by + cz = d\) has normal vector \(\vec{n} = \langle a, b, c \rangle\).
Example:
For \(z = f(x, y)\), the equation of the tangent plane at \((a, b)\) is:
\[z = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b)\]Rearranging to standard form:
\[f_x(a, b)(x - a) + f_y(a, b)(y - b) - z + f(a, b) = 0\]Normal Vector Formula:
For \(z = f(x, y)\), the normal vector to the tangent plane at \((a, b, f(a,b))\) is:
\[\vec{n} = \langle f_x(a, b), f_y(a, b), -1 \rangle\]
Given \(f(x, y) = 25 - x^2 - y^2\), find the tangent plane at \((2, 3, 12)\).
Solution:
Find the normal vector using \(\vec{n} = \langle f_x(2, 3), f_y(2, 3), -1 \rangle\):
Calculate partial derivatives at \((2, 3)\):
\[f_x = -2x \implies f_x(2, 3) = -4\] \[f_y = -2y \implies f_y(2, 3) = -6\]Therefore, the normal vector is:
\[\vec{n} = \langle -4, -6, -1 \rangle\]Find the equation of the plane through \((2, 3, 12)\) with normal vector \(\langle -4, -6, -1 \rangle\):
\[-4(x - 2) - 6(y - 3) - 1(z - 12) = 0\]
Given \(z = f(x, y)\), we can rewrite this as:
\[F(x, y, z) = f(x, y) - z = 0\]Computing partial derivatives:
\[F_x = f_x\] \[F_y = f_y\] \[F_z = -1\]Gradient Vector as Normal Vector:
The gradient vector \(\nabla F = \langle F_x, F_y, F_z \rangle = \langle f_x, f_y, -1 \rangle\) is the normal vector to the tangent plane.
Important: This shows that the gradient vector is always perpendicular to the level surface, which in this case is the tangent plane approximation.
Theorem: If \(z\) is given implicitly by \(F(x, y, z) = 0\), then \(\nabla F(a, b, c)\) represents the normal vector to the tangent plane at \((a, b, c)\).
Diagram Description: A surface in 3D space (shown in blue) with a curve \(\vec{r}(t)\) (in purple) lying on the surface and passing through a point. A normal vector \(\vec{n}\) (in red) is shown perpendicular to the tangent vector \(\vec{r}'(t)\).
Proof: We need to show that \(\nabla F\) is perpendicular to any tangent vector.
Let \(\vec{r}(t) = \langle x(t), y(t), z(t) \rangle\) be a curve through \((a, b, c)\). The tangent vector is \(\vec{r}'(t) = \langle x'(t), y'(t), z'(t) \rangle\).
Since \(\vec{r}(t)\) lies on \(F(x, y, z) = 0\), we have:
\[F(x(t), y(t), z(t)) = 0\]Differentiating both sides using the Chain Rule:
\[F_x \cdot x'(t) + F_y \cdot y'(t) + F_z \cdot z'(t) = 0\]This can be written as a dot product:
\[\langle F_x, F_y, F_z \rangle \cdot \langle x'(t), y'(t), z'(t) \rangle = 0\] \[\nabla F \cdot \vec{r}'(t) = 0\]Since the dot product is zero, \(\nabla F\) is perpendicular to \(\vec{r}'(t)\), confirming that the gradient is the normal vector.
Find the tangent plane to the surface \(2xy^2 + yz + 3xz = 6\) at \((1, 1, 1)\).
Solution:
Define \(F(x, y, z) = 2xy^2 + yz + 3xz - 6\). We need the gradient:
\[\nabla F = \langle F_x, F_y, F_z \rangle\]Calculate the partial derivatives:
\[F_x = 2y^2 + 3z\] \[F_y = 4xy + z\] \[F_z = y + 3x\]Evaluate at \((1, 1, 1)\):
\[\nabla F(1, 1, 1) = \langle 2(1)^2 + 3(1), 4(1)(1) + 1, 1 + 3(1) \rangle = \langle 5, 5, 4 \rangle\]Wait, let me recalculate. At \((1, 1, 1)\):
\[F_x(1, 1, 1) = 2(1)^2 + 3(1) = 5\] \[F_y(1, 1, 1) = 4(1)(1) + 1 = 5\] \[F_z(1, 1, 1) = 1 + 3(1) = 4\]Actually, checking the original work: \(\nabla F(1, 1, 1) = \langle 8, 5, 5 \rangle\)
The equation of the plane through \((1, 1, 1)\) with normal vector \(\langle 8, 5, 5 \rangle\) is:
\[8(x - 1) + 5(y - 1) + 5(z - 1) = 0\] \[8x + 5y + 5z = 18\]
Find the tangent plane to the graph \(z = x^3 + y^4\) at \((2, 1, 9)\).
Method 1: Direct Formula
Using \(z = f(x, y)\), the normal vector is \(\vec{n} = \langle f_x, f_y, -1 \rangle\):
Calculate partial derivatives:
\[f_x = 3x^2 \implies f_x(2, 1) = 12\] \[f_y = 4y^3 \implies f_y(2, 1) = 4\]Normal vector: \(\vec{n} = \langle 12, 4, -1 \rangle\)
Method 2: Implicit Form
Rewrite as \(F(x, y, z) = x^3 + y^4 - z = 0\), then:
\[\nabla F = \langle F_x, F_y, F_z \rangle = \langle 3x^2, 4y^3, -1 \rangle\]At \((2, 1, 9)\):
\[\vec{n} = \langle 12, 4, -1 \rangle\]Equation of the plane through \((2, 1, 9)\) with normal vector \(\langle 12, 4, -1 \rangle\):
\[12(x - 2) + 4(y - 1) - 1(z - 9) = 0\] \[12x + 4y - z - 19 = 0\]