Gradient and mean value theorem

So the two points are a little bit more than (1,3) and a little bit less than (1,3). Since (1,3) is on the line between (0,1) and (1,3), we can just say "s= (1,3).
  • #1
brad sue
281
0
Hi
please , can someone help me with this problem.
I need to know the procedure.

Let f(x,y)=x^3-xy. set a(0,1) and b(1,3).
Find a point c on the line sement[ab] for which
f(b)-f[a]= gradient(f[c]) * (b-a)


Thank you
B.
 
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  • #2
brad sue said:
Hi
please , can someone help me with this problem.
I need to know the procedure.
Let f(x,y)=x^3-xy. set a(0,1) and b(1,3).
Find a point c on the line sement[ab] for which
f(b)-f[a]= gradient(f[c]) * (b-a)

Thank you
B.
The "procedure" is to calculate everything you can there, then solve the equation for c. Since a= (0,1), f(a)= f(0,1)= 0^3- 0(1)= 0. Since b= (1, 3), f(b)= f(1,3)= 1^3- 1(3)= -2. f(b)- f(a)= -2- 0= -2.
I'm not particularly happy with the notation on the right side! a and b are points but you have to treat "b-a" as a vector subtraction: b- a= i+3j- (j)= i+ 2j. grad f= (3x- y)i- xj. Now that product "*" is a dot product:
grad f*b-a= ((3x-y)i- xj).(i+ 2j)= 3x-y- 2x= x- y= -2.
That's one equation for two unknown values, x and y. To get the other, use that fact that c= (x,y) is on the line between a and b. One way to do that is to write the line between a and b ((0,1) and (1,3)) as y= 2x+ 1 (do you see how I got that?) and solve the two equations x- y= 2 and y= 2x+ 1 for x and y. Another is to write parametric equations for the line: since b-a= i+ 3j, x= t, y= 2t+ 1. Replace x- y= 2 with those, solve for t and use that value of t to find x and y.
 
  • #3
Where did you find the solutions for this? I'm writing a paper and struggling to find info for use!
 
  • #4
What do you mean "where did you find the solutions for this?" I solved it myself!
 
  • #5
I guess I'm confused as to how you learned this. I couldn't find a lot of stuff online about it. As I said I'm working on a project involving this and I can't find anywhere to learn this stuff.
 
  • #6
I just copied what you wrote and I'll write comments in red of where I'm confused.

The "procedure" is to calculate everything you can there, then solve the equation for c. Since a= (0,1), f(a)= f(0,1)= 0^3- 0(1)= 0. Since b= (1, 3), f(b)= f(1,3)= 1^3- 1(3)= -2. f(b)- f(a)= -2- 0= -2.

I'm not particularly happy with the notation on the right side! a and b are points but you have to treat "b-a" as a vector subtraction: b- a= i+3j- (j)= i+ 2j. grad f= (3xshould this be squared?- y)i- xj. Now that product "*" is a dot product:
grad f*b-a= ((3x-y)i- xj).(i+ 2j)= 3x-y- 2x= x- y= -2.

How do you know that x-y=-2??

That's one equation for two unknown values, x and y. To get the other, use that fact that c= (x,y) is on the line between a and b. One way to do that is to write the line between a and b ((0,1) and (1,3)) as y= 2x+ 1 (do you see how I got that?) and solve the two equations x- y= 2

Why is x-y=2 now?

and y= 2x+ 1 for x and y. Another is to write parametric equations for the line: since b-a= i+ 3j, x= t, y= 2t+ 1. Replace x- y= 2 with those, solve for t and use that value of t to find x and y.
 
  • #7
kmac said:
I just copied what you wrote and I'll write comments in red of where I'm confused.

The "procedure" is to calculate everything you can there, then solve the equation for c. Since a= (0,1), f(a)= f(0,1)= 0^3- 0(1)= 0. Since b= (1, 3), f(b)= f(1,3)= 1^3- 1(3)= -2. f(b)- f(a)= -2- 0= -2.

I'm not particularly happy with the notation on the right side! a and b are points but you have to treat "b-a" as a vector subtraction: b- a= i+3j- (j)= i+ 2j. grad f= (3xshould this be squared?
Yes, that was, I hope, a typo on my part.

- y)i- xj. Now that product "*" is a dot product:
grad f*b-a= ((3x-y)i- xj).(i+ 2j)= 3x-y- 2x= x- y= -2.

How do you know that x-y=-2??
I just made a really dumb mistake! The derivative of [itex]x^3- xy[/itex] with respect to x is [itex]3x^2- y[/itex], not "3x- y"! I hope thoat was just a typo on my part and I am not really losing my mind.
Now [itex]grad f\cdot(b-a)= (3x^2- y)\vec{i}- x\vec{j})\cdot(\vec{i}+ 2\vec{j})[/itex][itex]= 3x^2- y- 2x[/itex] and that must be equal to -2: 3x^2- y- 2x= -2.

That's one equation for two unknown values, x and y. To get the other, use that fact that c= (x,y) is on the line between a and b. One way to do that is to write the line between a and b ((0,1) and (1,3)) as y= 2x+ 1 (do you see how I got that?) and solve the two equations x- y= 2

Why is x-y=2 now?
That was the mistaken equation I got before. It should be [itex]3x^2- y- 2x= -2[/itex]

and y= 2x+ 1 for x and y. Another is to write parametric equations for the line: since b-a= i+ 3j, x= t, y= 2t+ 1. Replace x- y= 2 with those, solve for t and use that value of t to find x and y.

So [itex]3x^2- y- 2x= 2[/itex] and y= 2x+1. Replace y in the first equation by 2x+1 and we have [itex]3x^2- 2x-1- 2x= -2[/itex] or [itex]3x^2- 4x+ 1= 0[/itex]. By the quadratic formula, the roots of that are
[tex]\frac{4\pm\sqrt{16- 9}}{6}= \frac{4\pm\sqrt{7}}{6}[/itex].

Since 7 is just a little less than 9, [itex]\sqrt{7}[/itex] is just a little less than 3. If we were to take the positive, we would have [itex]4+ \sqrt{7}[/itex] a little bit more than 7 and [itex](4+ \sqrt{7})/6[/itex] a little bit larger than 1. If we take the negative, we have [itex]4- \sqrt{7}[/itex] a little smaller than 1 and [itex](4-\sqrt{7})/6[/itex] between 0 and 1. In order that (x,y) be between (0,1) and (1,3) x must be between 0 and 1 and so [itex]x= (4- \sqrt{7})/6[/itex]. Then, of course, [itex]y= 2x+ 1= (4- \sqrt{7}}/3+ 1= (7- \sqrt{7})/3[/itex] which is a little larger than 1 and so between 1 and 3.
 

What is the gradient theorem?

The gradient theorem, also known as the fundamental theorem of calculus for line integrals, states that the line integral of a vector field over a curve can be calculated by evaluating the scalar field at the endpoints of the curve and taking the difference.

What is the mean value theorem?

The mean value theorem states that for a differentiable function on a closed interval, there exists a point within that interval where the slope of the tangent line is equal to the average rate of change of the function over the interval.

How is the gradient theorem related to the mean value theorem?

The gradient theorem is a special case of the mean value theorem, where the curve is a straight line and the scalar field is a constant function. This means that the slope of the tangent line at any point on the curve is equal to the constant value of the scalar field.

What are some real-life applications of the gradient and mean value theorems?

The gradient and mean value theorems are used in fields such as physics, engineering, economics, and statistics to calculate rates of change, find optimal solutions, and analyze data. For example, the gradient theorem is used in fluid dynamics to calculate fluid flow and in economics to determine the marginal rate of substitution. The mean value theorem is used in physics to analyze motion and in finance to calculate average rates of return.

What are some common misconceptions about the gradient and mean value theorems?

One common misconception is that the mean value theorem guarantees the existence of a maximum or minimum point on a function. However, this is not always the case as the function may not have a derivative at the point where the tangent line is horizontal. Another misconception is that the gradient theorem only applies to two-dimensional vector fields, when in fact it can be extended to higher dimensions. Additionally, some may mistakenly believe that the gradient theorem is only applicable to straight-line paths, when it can also be used for more complex curves.

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