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Section 6: Directional Derivatives and the Gradient Vector

Instructions

  • First, you should watch the concepts videos below explaining the topics in the section. 
  • Second, you should attempt to solve the exercises and then watch the videos explaining the exercises. 
  • Last, you should attempt to answer the self-assessment questions to determine how well you learned the material.
  • When you have finished the material below, you can start on the next section or return to the main several variable calculus page.

Concepts

  • The directional derivative
  • The gradient vector
 



If you would like to see more videos on this topic, click the following link and see the related videos. Note the related videos at the link are not required viewing.
Directional Derivatives and The Gradient Vector Conceptual V1


If you would like to see more videos on this topic, click the following link and see the related videos. Note the related videos at the link are not required viewing.
Directional Derivatives and The Gradient Vector Conceptual V2

Exercises


Directions: The following questions are an assessment of your understanding of the material above. If you are not sure of the answers, you may need to rewatch the videos.
  1. Find \(D_{{\bf{u}}}f(x,y)\) at the point \((1,2)\) in the direction of \(\left<1,-3\right>\) to the surface \(f(x,y)=x^3+2x^2y^2.\)

    \(D_{{\bf {u}}}f(1,2)=-\dfrac{5}{\sqrt{10}}\)


    If you would like to see more videos on this topic, click the following link and see the related videos. Note the related videos at the link are not required viewing.
    Directional Derivatives and The Gradient Vector Exercise V1


  2. Suppose \(f(x,y)=x^3-2xy+y^2\). Find \(D_{{\bf{u}}}f(x,y)\) at the point \((1,2)\) where \({\bf{u}}\) is the unit vector corresponding to \(\dfrac{\pi}{3}.\)

    \(D_{{\bf {u}}}f(1,2)=-\dfrac{1}{2}+\sqrt{3}\)


    If you would like to see more videos on this topic, click the following link and see the related videos. Note the related videos at the link are not required viewing.
    Directional Derivatives and The Gradient Vector Exercise V5


  3. If \(f(x,y,z)=z^3-x^2y\), find \(D_{{\bf{u}}}f(1,6,2)\) if \({\bf{u}}=\left<\dfrac{3}{13},\dfrac{4}{13},\dfrac{12}{13}\right>.\)

    \(D_{{\bf {u}}}f(1,6,2)=\dfrac{104}{13}\)


    If you would like to see more videos on this topic, click the following link and see the related videos. Note the related videos at the link are not required viewing.
    Directional Derivatives and The Gradient Vector Exercise V2


  4. Let \(f(x,y)=xe^y\)
    1. ​Find the rate of change of \(f\) at the point \((2,0)\) in the direction of the point \(P(2,0)\) to the point \(Q\left(\dfrac{1}{2},2\right).\)
    2. At the point \((2,0)\), in what direction does \(f\) have the maximum rate of change? What is the maximum rate of change?

      1. \(D_{{\bf{u}}}f(2,0)=1\)
      2. The direction of the maximum rate of change at the point \((2,0)\) is \(\nabla f(2,0)= \langle 1, 2\rangle\), and the maximum rate of change is \( \sqrt{5}.\)


      If you would like to see more videos on this topic, click the following link and see the related videos. Note the related videos at the link are not required viewing.
      Directional Derivatives and The Gradient Vector Exercise V3


  5. ​Find the maximum rate of change of \(f(x,y)=\tan(3x+2y)\) at the point \(\left(\displaystyle{{\pi}\over{6}}, -\displaystyle{{\pi}\over{8}}\right)\) and the direction in which it occurs.

    The direction of the maximum rate of change at \(\left(\dfrac{\pi}{6},-\dfrac{\pi}{8}\right)\) is \(\nabla f\left(\dfrac{\pi}{6},-\dfrac{\pi}{8}\right)=\langle6,4\rangle\), and the maximum rate of change is \(|\nabla f|=\sqrt{52}.\)


    If you would like to see more videos on this topic, click the following link and see the related videos. Note the related videos at the link are not required viewing.
    Directional Derivatives and The Gradient Vector Exercise V4


 

Self-Assessment Questions


Directions: The following self-assessment question are a measure of how well you understood the material in this section. 
  1. How is the geometric interpretation of the directional derivative of \(z=f(x,y)\) at a point \(P(x_0, y_0)\) in the direction of the unit vector \({\bf{u}}\) different from \(f_x(x_0, y_0)\) and \(f_y(x_0, y_0)\)?
  2. How can we use the directional derivative to find the steepest ascent or descent of a surface \(z=f(x,y)\)?
  3. What is the difference between the direction of the maximum rate of change of \(z=f(x,y)\) at a point \(P\) and the maximum rate of the change of \(z=f(x,y)\) at a point \(P\)?