Directional Derivative of a Function in a Given Direction

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SUMMARY

The directional derivative of the function \(\varphi = x^2 + \sin y - xz\) in the direction of the vector \( \mathbf{u} = \mathbf{i} + 2\mathbf{j} - 2\mathbf{k} \) at the point \((1, \frac{\pi}{2}, -3)\) is calculated using the normalized direction vector \(\hat{u} = \frac{1}{3}\mathbf{i} + \frac{2}{3}\mathbf{j} - \frac{2}{3}\mathbf{k}\) and the gradient \(\nabla\varphi = (2x - z)\mathbf{i} + \cos y \mathbf{j} - x\mathbf{k}\). At the specified point, the gradient evaluates to \((5, 0, -1)\). The final directional derivative is \(\frac{7}{3}\).

PREREQUISITES
  • Understanding of directional derivatives in multivariable calculus
  • Familiarity with gradient vectors and their properties
  • Knowledge of vector normalization techniques
  • Basic proficiency in trigonometric functions, specifically sine and cosine
NEXT STEPS
  • Study the concept of gradient vectors in multivariable calculus
  • Learn about the properties and applications of directional derivatives
  • Explore vector normalization methods in detail
  • Investigate the implications of trigonometric functions in calculus
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Students and professionals in mathematics, particularly those studying multivariable calculus, as well as educators looking to enhance their understanding of directional derivatives and gradient calculations.

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Homework Statement



Find the Directional derivative of [tex]\varphi[/tex] = x2+siny-xz in the direction of i+2j-2k at the point (1, [tex]\pi[/tex]/2 , -3)

The Attempt at a Solution



u = i+2j-2k

[tex]\left|[/tex]u[tex]\left|[/tex] = [tex]\sqrt{1^2+2^2+(-2)^2}[/tex] = 3


[tex]\Rightarrow[/tex] u[tex]\hat{}[/tex] = 1/3i+2/3j-2/3k

[tex]\nabla[/tex][tex]\varphi[/tex] = (2x-z)i+(Cos y)j-xk

at the point (1, [tex]\pi[/tex]/2 , -3)
u[tex]\hat{}[/tex][tex]\nabla[/tex][tex]\varphi[/tex] = 5/3 + 0 +2/3 = 7/3

i think i have this right?
 
Physics news on Phys.org
I think you have it right.
 
thanks a mill
 

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