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

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fluxer5
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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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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 .