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Section Gradient, Tangent Lines and Linear Approximations
Worksheet DC-3 Reading
Objectives
Things you should understand and know:
How tangent planes can be used to approximate a two-variable function at a point
The meaning of a directional derivative
The meaning of a gradient
Things you should be able to do:
Find the equation of a plane tangent to a given surface at a point
Use the tangent plane to approximate a function of two variables at a point
Use the differential to approximate the change in a function of two variables
Find the directional derivative in a given direction for a function of two variables
Find the gradient vector of a real-valued function
Use the gradient to find the tangent to a level curve of a function
Calculate the directional derivatives and gradients in three dimensions
You will be assessed on the following criteria:
Problem structure: Solution should be organized and easy to follow, with clear, complete steps that follow progressively from one to the next.
Tangent Planes: Understand the concept of tangent planes and their applications. Find the equation of a plane tangent to a given surface at a point. Use the tangent plane to approximate a function of two variables at a point.
Differentials: Use the differential to approximate the change in a function of two variables.
Directional Derivatives: Understand the concept of directional derivatives and their applications. Calculate the directional derivative of a function at a point in the direction of a unit vector.
Gradient: Understand the concept of the gradient and its applications. Calculate the gradient of a function. Find the directions of maximal/minimal rate of change of a function. Calculate the maximum/minimum rate of change. Find the equation of the tangent plane to a surface at a point.
All of the readings for this course come from the OpenStax Calculus Volume 3 textbook, which is available for free online at
https://openstax.org/details/books/calculus-volume-3 . You can read the textbook online or download a PDF version. The textbook is also available in print from various retailers if you prefer a physical copy.
For this lesson, you should read sections
4.4 Tangent Planes and Linear Approximation . You should read through the sections carefully, making sure to understand the definitions, examples, and key ideas. You should also work through the exercises at the end of each section to test your understanding and practice applying the concepts. Some of the exercises at the end of the section are assigned in the WP assignments.
Worksheet DC-3 Videos
Tangent Planes and Linear Approximation
Directional Derivatives and the Gradient
Worksheet DC-3 Written Practice
INFORMATION ABOUT ALL WRITTEN PRACTICE (WP) ASSIGNMENTS:
THE WP ASSIGNMENTS ARE DESIGNED TO PREPARE YOU FOR THE QUIZZES AND ASSESSMENTS
Work on every problem on every assignment.
If you get stuck or do not understand a solution, ask questions.
For each problem, try it once or twice before looking at solutions or asking for help.
Mark the problems you did correctly on the first try.
Highlight problems where you got help (using solutions, videos, a tutor, etc) -These are the problem types that you will need more practice on to be ready for the assessments.
Understand that just copying the solutions instead of working through the problems will greatly reduce your chances of success in this class.
Tangent Planes and Linear Approximations: Complete problems 171, 181, 196, 197, 203, 207
Directional Derivatives: Complete problems 263, 269, 281, 291, 307
Worksheet DC-3 Sample Outcomes
For the quiz on this outcome you would be given 50 minutes to complete 4 problems. Here are samples of the types of problems you will encounter:
1. Let
\(z=3x^2+y^2\text{.}\) Use the gradient to find the tangent plane at the point
\((1,-1,4)\text{.}\)
Hint .
To find the tangent plane, first rewrite the equation as
\(f(x,y,z) = 3x^2 + y^2 - z = 0\text{,}\) then compute the gradient of
\(f\) at the given point.
Answer .
The tangent plane is
\(6x - 2y - z = 4\text{.}\)
Solution .
First, rewrite the equation as
\(f(x,y,z) = 3x^2 + y^2 - z = 0\text{.}\)
Compute the gradient of
\(f\text{:}\)
\(\nabla f = \left(6x, 2y, -1\right)\)
Evaluate the gradient at the point
\((1,-1, 4)\text{:}\)
\(\nabla f(1,-1,4) = \left(6, -2, -1\right)\)
The equation of the tangent plane is:
\(6(x-1) - 2(y+1) - 1(z-4) = 0\)
\(6x - 6 - 2y - 2 - z + 4 = 0\)
2. Given the function
\(f(x,y)=\sqrt{20-x^2-7y^2}\)
(a) Find the linearization of
\(f\) at the point
\((2,1)\text{.}\)
Hint .
The linearization of
\(f\) at
\((a,b)\) is given by
\(L(x,y) = f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b)\text{.}\)
Answer .
\(L(x,y) = 3 + \frac{1}{3}(x-2) + \frac{7}{3}(y-1)\)
Solution .
First, find the partial derivatives of
\(f\text{:}\)
\(f_x = \frac{-x}{\sqrt{20-x^2-7y^2}}\)
\(f_y = \frac{-7y}{\sqrt{20-x^2-7y^2}}\)
Evaluate these at the point
\((2,1)\text{:}\)
\(f(2,1) = \sqrt{20-4-7} = \sqrt{9} = 3\)
\(f_x(2,1) = \frac{-2}{\sqrt{9}} = -\frac{2}{3}\)
\(f_y(2,1) = \frac{-7}{\sqrt{9}} = -\frac{7}{3}\)
\(L(x,y) = 3 - \frac{2}{3}(x-2) - \frac{7}{3}(y-1)\)
(b) Use the linearization to approximate
\(f(1.95,1.08)\text{.}\)
Hint .
To find the linear approximation, substitute the values into the linearization formula.
Answer .
\(f(1.95,1.08) \approx 3 + \frac{1}{3}(1.95-2) + \frac{7}{3}(1.08-1) = 3 - \frac{1}{6} + \frac{7}{3}(0.08) = 3 - \frac{1}{6} + \frac{7}{3} \cdot 0.08\)
Solution .
Substitute
\(x = 1.95\) and
\(y = 1.08\) into the linearization:
\(L(1.95,1.08) = 3 - \frac{2}{3}(1.95-2) - \frac{7}{3}(1.08-1)\)
\(L(1.95,1.08) = 3 - \frac{2}{3}(-0.05) - \frac{7}{3}(0.08)\)
\(L(1.95,1.08) = 3 + \frac{1}{30} - \frac{14}{75}\)
\(L(1.95,1.08) = \frac{91}{30} + \frac{14}{75}\)
\(L(1.95,1.08) = \frac{455}{150} - \frac{28}{150}\)
\(L(1.95,1.08) = \frac{427}{150} \approx 2.8467\)
3. Find the gradient of the function
\(f(x,y)=x^2\ln(y)\text{.}\)
Hint .
To find the gradient, compute the partial derivatives with respect to each variable.
Answer .
\(\nabla f = \left\langle 2x\ln(y), \frac{x^2}{y} \right\rangle\)
Solution .
First we compute the partial derivatives:
\(\frac{\partial f}{\partial x} = 2x\ln(y)\)
\(\frac{\partial f}{\partial y} = \frac{x^2}{y}\)
So,
\(\nabla f = \left\langle 2x\ln(y), \frac{x^2}{y} \right\rangle\text{.}\)
4. Given
\(f(x,y)=e^{-x}\sin y\text{,}\) find the directional derivative at the point
\((0, \frac{\pi}{3})\) in the direction of the vector
\(\vec{u} = \langle 3, -2 \rangle\text{.}\)
Hint .
To find the directional derivative, compute the gradient of
\(f\) and then take the dot product with the unit vector in the direction of
\(\vec{u}\text{.}\)
Answer . Solution .
First we compute the gradient:
\(\nabla f = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right\rangle\)
\(\frac{\partial f}{\partial x} = -e^{-x}\sin y\)
\(\frac{\partial f}{\partial y} = e^{-x}\cos y\)
So,
\(\nabla f = \left\langle -e^{-x}\sin y, e^{-x}\cos y \right\rangle\text{.}\)
Next, we evaluate the gradient at the point
\((0, \frac{\pi}{3})\text{:}\)
\(\nabla f(0, \frac{\pi}{3}) = \left\langle -e^{0}\sin \frac{\pi}{3}, e^{0}\cos \frac{\pi}{3} \right\rangle\)
\(\nabla f(0, \frac{\pi}{3}) = \left\langle -\frac{\sqrt{3}}{2}, \frac{1}{2} \right\rangle\text{.}\)
Now we find the unit vector in the direction of
\(\vec{u} = \langle 3, -2 \rangle\text{:}\)
\(\|\vec{u}\| = \sqrt{3^2 + (-2)^2} = \sqrt{13}\)
\(\hat{u} = \frac{1}{\sqrt{13}}\langle 3, -2 \rangle = \left\langle \frac{3}{\sqrt{13}}, \frac{-2}{\sqrt{13}} \right\rangle\)
Finally, we compute the directional derivative:
\(D_{\hat{u}}f(0, \frac{\pi}{3}) = \nabla f(0, \frac{\pi}{3}) \cdot \hat{u} = \left\langle -\frac{\sqrt{3}}{2}, \frac{1}{2} \right\rangle \cdot \left\langle \frac{3}{\sqrt{13}}, \frac{-2}{\sqrt{13}} \right\rangle\)
\(D_{\hat{u}}f(0, \frac{\pi}{3}) = \frac{-3\sqrt{39}-2\sqrt{13}}{26}\)
5. Given the surface
\(x-z=4\arctan(yz)\) and the point
\((1+\pi, 1, 1)\text{:}\)
(a) Find the equation of the tangent plane to the surface at the given point.
Hint .
To find the equation of the tangent plane, compute the gradient of the surface equation and evaluate it at the given point.
Answer .
tangent plane:
\(x - 2y - 3z + 2 = \pi - 4\)
Solution .
First we can rewrite the surface equation as
\(f(x,y,z) = x - z - 4\arctan(yz)\text{.}\) Then we compute the gradient:
\(\nabla F = \left\langle \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right\rangle\)
\(\frac{\partial F}{\partial x} = 1\)
\(\frac{\partial F}{\partial y} = -\frac{4z}{1 + (yz)^2}\)
\(\frac{\partial F}{\partial z} = -1 - \frac{4y}{1 + (yz)^2}\)
So,
\(\nabla F = \left\langle 1, -\frac{4z}{1 + (yz)^2}, -1 - \frac{4y}{1 + (yz)^2} \right\rangle\text{.}\)
Next, we evaluate the gradient at the point
\((1+\pi, 1, 1)\text{:}\)
\(\nabla F(1+\pi, 1, 1) = \left\langle 1, -\frac{4}{1 + 1}, -1 - \frac{4}{1 + 1} \right\rangle\)
\(\nabla F(1+\pi, 1, 1) = \left\langle 1, -2, -3 \right\rangle\text{.}\)
The equation of the tangent plane is:
\(1(x-(1+\pi)) - 2(y-1) - 3(z-1) = 0\)
\(x - (1+\pi) - 2(y-1) - 3(z-1) = 0\)
\(x - (1+\pi) - 2y + 2 - 3z + 3 = 0\)
\(x - 2y - 3z + 2 = \pi-4\)
(b) Find the equation of the normal line to the surface at the given point.
Hint .
To find the equation of the normal line, use the gradient of the surface equation evaluated at the given point as the direction vector.
Answer .
normal line:
\(\frac{x-(1+\pi)}{1} = \frac{y-1}{-2} = \frac{z-1}{-3}\)
Solution .
Since the gradient at the given point is
\(\nabla F(1+\pi, 1, 1) = \left\langle 1, -2, -3 \right\rangle\text{,}\) the equation of the normal line is:
\(\frac{x-(1+\pi)}{1} = \frac{y-1}{-2} = \frac{z-1}{-3}\)
