Find the gradient of a function at a given point, sketch level curve

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SUMMARY

The discussion focuses on finding the gradient of the function \(f(x,y)=\sqrt{2x+3y}\) at the point (-1,2), which is calculated as \(\frac{1}{2}\hat{i}+\frac{3}{4}\hat{j}\). Additionally, it addresses the sketching of the level curve associated with this gradient. The level curve equation is derived from the function value at the point, resulting in \(4=2x+3y\), which corresponds to \(f(-1,2)=2\). This establishes the relationship between the function value and the level curve.

PREREQUISITES
  • Understanding of gradient vectors in multivariable calculus
  • Familiarity with level curves and their significance
  • Basic knowledge of the square root function and its properties
  • Ability to perform partial derivatives
NEXT STEPS
  • Study the properties of gradient vectors in multivariable calculus
  • Learn how to derive level curves from multivariable functions
  • Explore the implications of gradients in optimization problems
  • Practice sketching level curves for various functions
USEFUL FOR

Students in calculus courses, educators teaching multivariable calculus, and anyone interested in understanding gradients and level curves in mathematical functions.

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So I have a function \(f(x,y)=\sqrt{2x+3y}\) and need to find the gradient at the point (-1,2). I got this part already, its \(\frac{1}{2}\hat{i}+\frac{3}{4}\hat{j}\). The part I'm having trouble with is when it asks me to sketch the gradient with the level curve that passes through (-1,2).
The back of the book has the answer and it calls the line passing through the point given point \(4=2x+3y\). I know that in relation to the given function this would also equal \(2=\sqrt{2x+3y}\). So where does that 2 come from?
 
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Re: find the gradient of a function at a given point, sketch level curve

The 2 comes from:

$f(-1,2)=\sqrt{2(-1)+3(2)}=\sqrt{4}=2$
 

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