Directional Derivatives: the maximum rate of change of f.

[DavidAp]

The maximum rate of change of f at the given point and the direction in which it occurs.
f(x,y)=(y^2)/x, (2,4)

Answer: 4√2, <-1,1>
_________________________

For this problem, since I couldn't find the upside down triangle, I am going to use delta to represent the gradient of a function.

f(x,y) = (y^2)/x

Δf(x,y)
= <-(y^2)/(x^2), (2yx)/(x^2)>
=<-(y^2)/(x^2), (2y)/(x)>

Δf(2,4)
=<-(4^2)/(2^2), (2*4)/(2)>
=<-16/4, 4>
=<-4,4> and since Δf is a vector i can divide this by the scalar 4 to get,
=<-1,1>

However, I have no idea on how to approach the problem when it comes to finding the maximum rate of change (which, in this problem, is 4√2). I tried finding the unit vector of Δf and multiplying it by Δf in hopes that it will give me the maximum rate of change but it didn't. Also, I found that this approach is the wrong approach, not only because it gave me the wrong answer but, because the answer this would give me would depend on whether or not I changed Δf by the scalar 4.

Can somebody guide me in the right direction! I found that the direction of the maximum rate of change is the gradient of the function but to find that rate of change vexes me!

Thank you for taking the time to review my problem.

 

[SammyS]

Have a "nabla" ▽

 

[Dick]

You are doing ok so far. The gradient is <-4,4>. The maximum rate of change is the dot product of a unit vector pointing in the direction of <-4,4> with <-4,4>. What's the unit vector? What's the dot product?

 

[DavidAp]

You are doing ok so far. The gradient is <-4,4>. The maximum rate of change is the dot product of a unit vector pointing in the direction of <-4,4> with <-4,4>. What's the unit vector? What's the dot product?

The maximum rate of change would be the direction derivative, (thank you SammyS for the "nabla!") Du▽f(2,4) = <-4,4>*u

Does it matter if it's <-4,4> or <-1,1> since they are the same vector? What doing you mean pointing in the direction of <-4,4> with <-4,4>? Isn't that a zero vector which is basically a dot?

Give me another hint! How do I find u! (Haha)

 

[Dick]

The maximum rate of change would be the direction derivative, (thank you SammyS for the "nabla!") Du▽f(2,4) = <-4,4>*u

Does it matter if it's <-4,4> or <-1,1> since they are the same vector? What doing you mean pointing in the direction of <-4,4> with <-4,4>? Isn't that a zero vector which is basically a dot?

Give me another hint! How do I find u! (Haha)

The gradient is <-4,4>. Not <-1,1>. They aren't the same vector. You can't change that. What's a unit vector in the direction of <-4,4>? Answer that first. Then take the dot product of it with <-4,4>.

 

[DavidAp]

The gradient is <-4,4>. Not <-1,1>. They aren't the same vector. You can't change that. What's a unit vector in the direction of <-4,4>? Answer that first. Then take the dot product of it with <-4,4>.

"I tried finding the unit vector of Δf and multiplying it by Δf in hopes that it will give me the maximum rate of change but it didn't" :/

u
= <-4,4>/sqrt((-4)^2 + 4^2)
= <-4,4>/sqrt(16+16)
= <-4,4>/sqrt(16)sqrt(2)
= <-4,4>/4sqrt(2)
=<-1/√2, 1/√2>

which you would also get if you changed the vector <-4,4> to <-1,1>.

Du▽f(2,4)
= <-4,4><-1/√2, 1/√2>
= 4/√2 + 4/√2
= 8/√2

However, the answer is 4√2. What did I do wrong?! Is that not how you find the unit vector?

 

[Dick]

"I tried finding the unit vector of Δf and multiplying it by Δf in hopes that it will give me the maximum rate of change but it didn't" :/

u
= <-4,4>/sqrt((-4)^2 + 4^2)
= <-4,4>/sqrt(16+16)
= <-4,4>/sqrt(16)sqrt(2)
= <-4,4>/4sqrt(2)
=<-1/√2, 1/√2>

which you would also get if you changed the vector <-4,4> to <-1,1>.

Du▽f(2,4)
= <-4,4><-1/√2, 1/√2>
= 4/√2 + 4/√2
= 8/√2

However, the answer is 4√2. What did I do wrong?! Is that not how you find the unit vector?

All is good. 8/sqrt(2) is the same as 4*sqrt(2). Isn't it?

 

[DavidAp]

All is good. 8/sqrt(2) is the same as 4*sqrt(2). Isn't it?

*checks with WolfraAlpha*
It... it is! >< I had the right answer the whole time! I swear, I check that in my calculator earlier and it said they weren't the same... must have plugged it in incorrectly.

Oh man, I feel silly. This entire halt in my life to solve this one problem and I had it right the whole time!

Thank you for your patience! I greatly appreciate it!

 

[Dick]

*checks with WolfraAlpha*
It... it is! >< I had the right answer the whole time! I swear, I check that in my calculator earlier and it said they weren't the same... must have plugged it in incorrectly.

Oh man, I feel silly. This entire halt in my life to solve this one problem and I had it right the whole time!

Thank you for your patience! I greatly appreciate it!

Don't worry. We've all had the same lack of faith in our correctness. It's good to doubt.