Maximizing Growth Direction from a Point on a Surface

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SUMMARY

The discussion focuses on maximizing growth direction from a point on a surface defined by the function f(x,y,z) = (x^2)y - x(e^z) at the point Po = (2, -1, π). Key tasks include finding the gradient at Po and the rate of change of f in the direction of the vector u = i - 2j + k. The participants clarify that the angle for maximum growth corresponds to the directional derivative and the gradient, specifically when the angle between the unit vector and the gradient is zero. The directional derivative is expressed as D_θ f(x,y) = (cos θ i + sin θ j) · (∂f/∂x i + ∂f/∂y j), with maximum growth occurring when the derivative with respect to θ equals zero.

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  • Understanding of multivariable calculus, specifically gradients and directional derivatives.
  • Familiarity with vector operations, including dot products and unit vectors.
  • Knowledge of the function notation and partial derivatives.
  • Basic comprehension of angles in relation to vector directions.
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  • Study the concept of directional derivatives in multivariable calculus.
  • Learn how to compute gradients for functions of multiple variables.
  • Explore the relationship between gradients and tangent planes to surfaces.
  • Investigate the implications of angles in vector calculus, particularly in optimization problems.
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Students and professionals in mathematics, physics, and engineering who are working with multivariable functions and seeking to optimize growth directions on surfaces.

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I have f(x,y,z)=(x^2)y-x(e^z) and point Po=(2,-1,pi)
I need to find
a) gradient at point Po ( done)
b) Rate of change of f at point Po in the direction of vector u=i-2j+k (it's also done)
c) Unit vector in the direction of fastest growth of f at Po.

I can't find formulas for a last on. Does it come from a) and b)?
I know that angle should be zero but I am not sure what angle it is.
 
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At what angle is the growth greatest? Its got to do with the directional derivative and the gradient.
 
The growth is greatest when angle equals zero.
 
When WHAT angle equals zero?

(You are almost there!)
 
I think it's the angle between unit vector and tangent vector to f at Po.
If I'm correct, I'm not sure how should I put it on a paper.
 
Youre right. When you take a directional derivative what kind of product do you use (for the two vectors). One of these two is at a maximum when the angle is zero, that's the one you want to use.

If you're still stuck, show the defining expression for a directional derivative.
 
Last edited:
Not "tangent vector" because you are not talking about a curve.

The gradient is the "derivative" of a function of 2 or more variables. The derivative of f(x,y) in the direction of angle \theta if given by
D_\theta f(x,y)= (cos \theta i+ sin \theta j)\dot(\frac{\partial f}{\partial x}i+ \frac{\partial f}{\partial y})
That will have a maximum (with respect to \theta) where it's derivative with respect to \theta equals 0:
-cos\theta \frac{\partial f}{\partial x}+ sin\theta\frac{\partial f}{\partial y}= 0
That means that
tan\theta= \frac{sin\theta}{cos\theta}= \frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}

Think about what that means in terms of the components of the gradient of f.
 
It is tangent to the surface at that point and I think that is what he is meaning, but when youre dotting D_{\theta}f(x,y) why did we get a negative cosine?

You also forgot your j unit vector in the first expression.
 
Yes, my mistake I thought of tangent vector to a surface.
Why should I dot f(x,y) if I have f(x,y,z)?
Still confused.
 

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