Find dV for V= 200/(x2 + y2 )1/2 when x=2 and y=1 - Homework Help

I think this is right, but it's always worth checking.) So, find the derivative of V with respect to x, along this line.In summary, the rate of change of the electric potential in a region of space can be described by the equation V = 200/(x2 + y2)1/2. To find the strength and direction of the electric field at a specific point (x,y), the equation E = -dV/ds can be used, where s represents a distance. However, since only one point is given, the directional derivative must be used instead. This can be found by taking the derivative of V with respect to x, along the line y = (1/2)x.
  • #1
IslandHead
10
0

Homework Statement

The electrical potential can be described by the following equation:
V= 200/(x2 + y2 )1/2 find dV when x=2 and y=1


Homework Equations


n/a


The Attempt at a Solution


dv/d(x,y) = ∂/∂x + ∂/∂y
=200/(x2 + y2)(1/2) +200/(x2 + y2)(1/2)
replace variables with C where necessary and rearrange for easy of deriving =200(x2 + C)(1/2) +200(C + y2)(1/2)
Chain rule and exponent rule
=200(-1/2)(2x)(x2 + C)(-3/2) +200(-1/2)(2y)(C + y2)(-3/2)
=-200x(x2 + C)(-3/2) + -200y)(C + y2)(-3/2)
=-200x(x2 + y2) )(-3/2) + -200y)(x2+ y2)(-3/2)
Then I plug in the values x=2 and y=1 and dV=-53.7 V/m, I'm not very confident with multivariable calculus and just wanted to check to see if this is right.I've never had to do this before, it is fir physics, not a math class
 
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  • #2
IslandHead said:

Homework Statement

The electrical potential can be described by the following equation:
V= 200/(x2 + y2 )1/2 find dV when x=2 and y=1

Homework Equations


n/a

The Attempt at a Solution


dv/d(x,y) = ∂/∂x + ∂/∂y
=200/(x2 + y2)(1/2) +200/(x2 + y2)(1/2)
replace variables with C where necessary and rearrange for easy of deriving =200(x2 + C)(1/2) +200(C + y2)(1/2)
Chain rule and exponent rule
=200(-1/2)(2x)(x2 + C)(-3/2) +200(-1/2)(2y)(C + y2)(-3/2)
=-200x(x2 + C)(-3/2) + -200y)(C + y2)(-3/2)
=-200x(x2 + y2) )(-3/2) + -200y)(x2+ y2)(-3/2)
Then I plug in the values x=2 and y=1 and dV=-53.7 V/m, I'm not very confident with multivariable calculus and just wanted to check to see if this is right.I've never had to do this before, it is fir physics, not a math class

Your calculation looks OK, but the notation is a little off. It should be like this:
dV = -200x(x2 + y2) (-3/2)*dx - 200y(x2+ y2)(-3/2)*dy

To calculate dV, you need to know x and y, but you also need dx and dy, which you don't show. Did you forget to include that information?
 
  • #3
Mark44 said:
Your calculation looks OK, but the notation is a little off. It should be like this:
dV = -200x(x2 + y2) (-3/2)*dx - 200y(x2+ y2)(-3/2)*dy

To calculate dV, you need to know x and y, but you also need dx and dy, which you don't show. Did you forget to include that information?
I am not very familiar with the notation. I haven't taken any multivariable calculus, and I'm a little rusty in my calculus in general and am just trying to solve a problem for my first year physics course. Do I need to solve for dx and dy, or do I just leave the dx and dy as bits of notation? I thought dx=-200x(x2 + y2) (-3/2) and dy=- 200y(x2+ y2)(-3/2)
 
  • #4
IslandHead said:
I am not very familiar with the notation. I haven't taken any multivariable calculus, and I'm a little rusty in my calculus in general and am just trying to solve a problem for my first year physics course. Do I need to solve for dx and dy
No, this should be given information.
IslandHead said:
, or do I just leave the dx and dy as bits of notation?
If you're not given values for dx and dy (or Δx and Δy), you'll have to leave these symbols in, and you won't be able to get a numeric value for dV. Are you sure there isn't some more information, something along the lines of x changes by ... and y changes by ... ? These would be your dx and dy.
IslandHead said:
I thought dx=-200x(x2 + y2) (-3/2) and dy=- 200y(x2+ y2)(-3/2)

No, what you have as dx is the partial of V with respect to x, ##\frac{\partial V}{\partial x}##.
And what you have as dy is the partial of V with respect to y, ##\frac{\partial V}{\partial y}##.

The full equation is ##dV = \frac{\partial V}{\partial x}~dx + \frac{\partial V}{\partial y}~dy ##
 
  • #5
The question is as follows:
"The electric potential in a region of space is V= 200/(x2 + y2 )1/2 , where x and y are in meters. What are the strength and direction of the electric field at (x,y) = (2.0m, 1.0m)?"
the relevant equation from the chapter is E=-dv/ds = the negative slope potential graph. X and Y aren't changing, I'm just trying to find the negative slope of V at (2.0m, 1.0m)
 
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  • #6
Well, that's very different from your original question. dV/ds is the rate of change of voltage with respect to s, but I don't know what s is, other than possibly distance along some curve or line.
 
  • #7
Mark44 said:
Well, that's very different from your original question. dV/ds is the rate of change of voltage with respect to s, but I don't know what s is, other than possibly distance along some curve or line.
s is a distance, in the problem there is only one point given though... So in this scenario it would be -dV/d(x,y) cause s would be defined by the x,y coordinate system.
 
  • #8
IslandHead said:
s is a distance, in the problem there is only one point given though... So in this scenario it would be -dV/d(x,y)
No, I'm pretty sure there's no such thing. You can't take a derivative with respect to two variables.
IslandHead said:
cause s would be defined by the x,y coordinate system.

My guess is that you need to take a directional derivative to find the rate of change of V in the direction of (2, 1). Along this line, y = (1/2)x.
 

FAQ: Find dV for V= 200/(x2 + y2 )1/2 when x=2 and y=1 - Homework Help

1. What is an implicit derivative?

An implicit derivative is a type of derivative that is used to find the rate of change of a function where the dependent variable is not explicitly expressed in terms of the independent variable. This means that the function is not in the form of y = f(x), but rather in a more complex form such as x^2 + y^2 = 25.

2. How is an implicit derivative calculated?

An implicit derivative is calculated using the chain rule of differentiation. This involves treating the dependent variable as a function of the independent variable and taking the derivative of both sides of the equation. The resulting expression will be the implicit derivative.

3. Why is an implicit derivative useful?

An implicit derivative is useful because it allows us to find the rate of change of a function even when it is not in a simple form. This is important in many fields of science, such as physics and engineering, where functions are often expressed implicitly.

4. What are some common applications of implicit derivatives?

Implicit derivatives have many applications in science and engineering. They are used to calculate rates of change in complex systems, such as the velocity of an object moving along a curved path or the growth rate of a population. They are also used in optimization problems to find the maximum or minimum value of a function.

5. Are there any tips for solving implicit derivative problems?

When solving implicit derivative problems, it is important to carefully apply the chain rule and to be familiar with basic differentiation rules. It can also be helpful to rewrite the equation in a simpler form before taking the derivative. Practice and familiarity with implicit derivatives will also improve problem-solving skills.

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