I need to find an expression for the path moved along going always in the steepest direction up a hill whose height is given by f(x,y)=x^2y-2xy+5, starting at (2,1). There is a hint that says to express the path first as parametric equations but I don't know how. Can anyone help?
Multivariable Calc.: steepest path along f(x,y)=x^2y-2xy+5
- Thread starter nivekious
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nivekious said:
I feel so sorry for you, that no one reply your question. Too bad it's too late for me to tell you the answer but just for people to looked at your post and don't know the answer, there it is.
Note that a theore says that the steepest direction to go uphill is in the direction of the gradien of the function steepest direction to go down hill it's in the opposite direction of the gradien. See attachement for more info.
Next time try google it.
Attachments
The gradient gives the direction of steepest ascent.
What they mean by parametrize the curve is to let the path be represented by the position function
\(\displaystyle r(t)=x(t)+y(t)\)
A tangent vector at each point (x(t), y(t)) is given by
\(\displaystyle r'(x)=\frac{dx}{dt}i+\frac{dy}{dy}j\)
Because we seek the steepest path, the directions of r'(t) and \(\displaystyle {\nabla}f(x,y)\) are the same at each point on the path.
\(\displaystyle f_{x}=2xy-2y\)
\(\displaystyle f_{y}=x^{2}-2x\)
Now, can you finish up?.
What they mean by parametrize the curve is to let the path be represented by the position function
\(\displaystyle r(t)=x(t)+y(t)\)
A tangent vector at each point (x(t), y(t)) is given by
\(\displaystyle r'(x)=\frac{dx}{dt}i+\frac{dy}{dy}j\)
Because we seek the steepest path, the directions of r'(t) and \(\displaystyle {\nabla}f(x,y)\) are the same at each point on the path.
\(\displaystyle f_{x}=2xy-2y\)
\(\displaystyle f_{y}=x^{2}-2x\)
Now, can you finish up?.
galactus said:
OMG, the person posted the question in 2008! and you think he/she still need to know specifically how to finish the question? it's really really really past due already! the answer to this assignment is probobly already known by this person. Come on!