Partial Derivative: Finding the vector on a scalar field at point (3,5)

In summary, the problem involves finding the component of del ∅ at the point (3,5) that makes a -60o angle with the x-axis, which can be done by taking the dot product of del ∅ and a unit vector that makes a -60o angle with the x-axis. The answer is -13.02.
  • #1
fluxer5
1
0

Homework Statement


A scalar field is given by the function: ∅ = 3x2y + 4y2
a) Find del ∅ at the point (3,5)
b) Find the component of del ∅ that makes a -60o angle with the axis at the point (3,5)

Homework Equations


del ∅ = d∅/dx + d∅/dy

The Attempt at a Solution


I completed part a:
del ∅ = (6xy+4y2)[itex]\hat{i}[/itex] + (3x2+8y)[itex]\hat{j}[/itex] = 120[itex]\hat{i}[/itex] + 67[itex]\hat{j}[/itex] (this answer is correct)

I am having trouble with part b. My gut feeling is that I need to take a dot product; project the vector from 'part a' onto the vector that makes "a -60o angle with the axis."

Assuming the equation is |a||b|cosθ,
  • I think a = 120[itex]\hat{i}[/itex] + 67[itex]\hat{j}[/itex]
  • I'm not sure what b is. Maybe a unit vector?
  • I'm thinking θ is 60o + atan(67/120)

Also, I'm assuming the "axis" that the problem refers to is the x-axis. The answer is -13.02.

Thank you for any help.

EDIT: Oops. 120i + 67j is not correct; it's 90i + 67j. |90i + 67j|cos(60+atan(67/90)) = -13.02
 
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  • #2
fluxer5 said:

Homework Statement


A scalar field is given by the function: ∅ = 3x2y + 4y2
a) Find del ∅ at the point (3,5)
b) Find the component of del ∅ that makes a -60o angle with the axis at the point (3,5)

Homework Equations


del ∅ = d∅/dx + d∅/dy

The Attempt at a Solution


I completed part a:
del ∅ = (6xy+4y2)[itex]\hat{i}[/itex] + (3x2+8y)[itex]\hat{j}[/itex] = 120[itex]\hat{i}[/itex] + 67[itex]\hat{j}[/itex] (this answer is correct)

I am having trouble with part b. My gut feeling is that I need to take a dot product; project the vector from 'part a' onto the vector that makes "a -60o angle with the axis."

Assuming the equation is |a||b|cosθ,
  • I think a = 120[itex]\hat{i}[/itex] + 67[itex]\hat{j}[/itex]
  • I'm not sure what b is. Maybe a unit vector?
  • I'm thinking θ is 60o + atan(67/120)

Also, I'm assuming the "axis" that the problem refers to is the x-axis. The answer is -13.02.

Thank you for any help.

EDIT: Oops. 120i + 67j is not correct; it's 90i + 67j. |90i + 67j|cos(60+atan(67/90)) = -13.02
Hello luxer5. Welcome to PF !

[itex]\cos(\theta)\hat{i}+\sin(\theta)\hat{j}[/itex] is a vector that makes an angle of θ with the positive x-axis. It also happens to have a magnitude of 1 .
 
1.

What is a partial derivative?

A partial derivative is a mathematical concept used in multivariable calculus to find the rate of change of a function with respect to one of its variables, while holding all other variables constant. It is denoted by ∂ (the partial symbol) and the variable with respect to which we are taking the derivative.

2.

How is a partial derivative different from a regular derivative?

A partial derivative is different from a regular derivative in that it involves taking the derivative of a multivariable function with respect to only one of its variables, while treating all other variables as constants. This is because in a multivariable function, the rate of change can vary depending on which variable we are looking at, so we need to specify which variable we are taking the derivative with respect to.

3.

What is a scalar field?

A scalar field is a mathematical concept that assigns a scalar value (such as temperature, pressure, or density) to every point in space. In other words, it is a function that takes in a point in space as an input and gives a single numerical value as an output.

4.

How do you find the vector on a scalar field at a specific point?

To find the vector on a scalar field at a specific point, we first need to take the partial derivative of the scalar field with respect to each variable. This will give us the components of the vector. Then, we can plug in the coordinates of the specific point into these components to get the vector at that point.

5.

What information can we get from finding the vector on a scalar field at a point?

By finding the vector on a scalar field at a specific point, we can determine the direction and magnitude of the rate of change of the scalar field at that point. This can be useful in various applications, such as determining the direction of heat flow in a material or the direction of fluid flow in a fluid dynamics problem.

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