Pauls Online Notes
Pauls Online Notes
Home / Calculus III / Partial Derivatives / Directional Derivatives
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 2-7 : Directional Derivatives

For problems 1 & 2 determine the gradient of the given function.

  1. \(\displaystyle f\left( {x,y} \right) = {x^2}\sec \left( {3x} \right) - \frac{{{x^2}}}{{{y^3}}}\) Solution
  2. \(f\left( {x,y,z} \right) = x\cos \left( {xy} \right) + {z^2}{y^4} - 7xz\) Solution

For problems 3 & 4 determine \({D_{\vec u}}f\) for the given function in the indicated direction.

  1. \(\displaystyle f\left( {x,y} \right) = \cos \left( {\frac{x}{y}} \right)\) in the direction of \(\vec v = \left\langle {3, - 4} \right\rangle \) Solution
  2. \(f\left( {x,y,z} \right) = {x^2}{y^3} - 4xz\) in the direction of \(\vec v = \left\langle { - 1,2,0} \right\rangle \) Solution
  3. Determine \({D_{\vec u}}f\left( {3, - 1,0} \right)\) for \(f\left( {x,y,z} \right) = 4x - {y^2}{{\bf{e}}^{3x\,z}}\) direction of \(\vec v = \left\langle { - 1,4,2} \right\rangle \). Solution

For problems 6 & 7 find the maximum rate of change of the function at the indicated point and the direction in which this maximum rate of change occurs.

  1. \(f\left( {x,y} \right) = \sqrt {{x^2} + {y^4}} \) at \(\left( { - 2,3} \right)\) Solution
  2. \(f\left( {x,y,z} \right) = {{\bf{e}}^{2x}}\cos \left( {y - 2z} \right)\) at \(\left( {4, - 2,0} \right)\) Solution