Multivariable calculus problem involving partial derivatives along a surface

  • #1
sss1
50
2
Homework Statement
Image
Relevant Equations
NA
I just wanted to know if my solution to part (b) is correct. Here's what I did:
I took the partial derivative with respect to x and y, which gave me
{font:{family:Arial,color:#000000,size:11},aid:null,backgroundColor:#ffffff,id:11,code:$$\\frac{-x}{\\left(x^{2}+y^{2}\\right)^{3/2}},\\,\\frac{-y}{\\left(x^{2}+y^{2}\\right)^{3/2}}$$,type:$$,ts:1695899323324,cs:mLppUJ1pren6kF2+GAN3Fw==,size:{width:192,height:40}}
respectively.
Then I computed the partial derivatives at (-3,4) which gave me 3/125 for partial derivative wrt x and -4/125 for partial derivative wrt y
Then since directional derivative requires a direction, I just chose an arbitrary one, uhat=(a,b)
since u is a unit vector that means sqrt(a^2+b^2)=1, or a^2+b^2=1.
I then solved for a, which is the plus minus of sqrt(1-b^2).

I just chose to use the positive answer here instead.
So the directional derivative is 3sqrt(1-b^2)/125-4b/125
To maximise this I took the derivative, which is -3b/125sqrt(1-b^2)-4/125
I set it to 0 and solved for b, which gave me -4/5.
So there is a local maximum for the directional derivative at b=-4/5 (I evaluated the second derivative and it was positive).
So that means on both sides of b=-4/5 the directional derivative decreases.
Subsituting b=-0.8 into my formula for a, a=sqrt(1-b^2), gives me a=0.6
So the directional derivative should be a maximum in the direction given by the unit vector (0.6, -0.8, 0) with magnitude 0.6(3/125)-0.8(-4/125) which is 0.04?

Although I didn't try the negative answer for a, a=-sqrt(1-b^2), I believe that this will yield a smaller answer because if a is negative, then a negative number times a positive number (3/125) will decrease the answer overall? Is my logic correct?
Screen Shot 2023-09-28 at 20.43.41.png
 
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  • #2
The gradient of a function evaluated at a point tells you the direction in which the function increases most rapidly at that point. This would give you the answer more quickly.
 
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  • #3
PeroK said:
The gradient of a function evaluated at a point tells you the direction in which the function increases most rapidly at that point. This would give you the answer more quickly.
Can you explain how to do it using this method?
 
  • #4
sss1 said:
Can you explain how to do it using this method?
You could have just plugged ##x = -3, y = 4## into the derivative you calculated.
 
  • #5
PeroK said:
You could have just plugged ##x = -3, y = 4## into the derivative you calculated.
derivative as in df/dt?
 
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  • #6
sss1 said:
derivative as in df/dt?
Look up what gradient means for a multi-variable function.
 
  • #7
PeroK said:
Look up what gradient means for a multi-variable function.
f_x(x,y)i+f_y(x,y)j? I plugged in (-3,4) and it gave me (3/125, -4/125)
 
  • #8
sss1 said:
f_x(x,y)i+f_y(x,y)j? I plugged in (-3,4) and it gave me (3/125, -4/125)
But isn't that the gradient at (-3,4)? How is that the direction f increases most rapidly?
 

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