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Consider f and c below. f(x, y) = x2 i + y2 j c is the arc of the parabola y = 4x2 from (−1, 4) to (0, 0) (a) find a function f such that f = ∇f. f(x, y) = (b) use part (a) to evaluate c ∇f · dr along the given curve
c.

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LammettHash
We're looking for a scalar function [tex]f(x,y)[/tex] such that

[tex]\nabla f(x,y)=\dfrac{\partial f}{\partial x}\,\mathbf i+\dfrac{\partial f}{\partial y}\,\mathbf j=x^2\,\mathbf i+y^2\,\mathbf j=\mathbf f(x,y)[/tex]

We have

[tex]\dfrac{\partial f}{\partial x}=x^2\implies f(x,y)=\dfrac{x^3}3+g(y)[/tex]
[tex]\implies\dfrac{\partial f}{\partial y}=\dfrac\partial{\partial y}\left[\dfrac{x^3}3+g(y)\right][/tex]
[tex]\implies y^2=\dfrac{\mathrm dg}{\mathrm dy}[/tex]
[tex]\implies g(y)=\dfrac{y^3}+C[/tex]

So

[tex]f(x,y)=\dfrac{x^3+y^3}3+C[/tex]

By the fundamental theorem of calculus (for line integrals), we then have

[tex]\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\int_{\mathcal C}\nabla f\cdot\mathrm d\mathbf r=f(0,0)-f(-1,4)=0-21=-21[/tex]

To answer this question, we need to make use of the concept of Potential Function (U)

f = grad U

The solution is:

U = x³/3  +  y³/3 +  K           f = grad×U

b)  - 54.41

f( x,y ) = x²×i + y²× j    is a vectorial field that could be obtained from the gradient of  an scalar, then

f( x, y ) = grad. U if that is true then:

dU/dx = x²       ⇒    dU = x²× dx   ⇒ Uₓ = x³/3 + g(y)          

dU/dy = y²      ⇒     according to that

g´(y) = y²        ⇒   ∫g´(y) = y³/3 + K    plugging that value in Uₓ we get the potential function U, or a family of functions according to  constant (K)

U = x³/3  +  y³/3 +  K

b) To calculate  ∫ grad.F * dr

C is the parabola     y = 4×x²

we parametrize the curve as:

x = t         then    y = 4×t²          r = ( t , 4×t² )        P ( -1 , 4)  to Q ( 0 , 0)                            for        - 1 ≤ t ≤ 0

If  x = t     dx = dt     and   dy = 8×t×dt        so       dr =  ( 1 + 8×t ) dt

We choose K = 0 then U = x³/3 + y³/3

then   U = ( t³/3 + 64×t⁶ )

∫₋₁⁰  ( t³/3 + 64×t⁶) ×  ( 1 + 8×t ) dt  = ∫ [t³/3 + (8/3)×t⁴ +  64×t⁶ + 512×t⁷ ] ×dt

= t⁴/12 + (8/15)× t⁵  + (64/7) ×t⁷ + 512 /8× t⁸

= t⁴/12 + (8/15)× t⁵  + (64/7) ×t⁷ + 64 ×t⁸ | ₋₁⁰

= - [ 1/12 - 8/15 - 64/7 + 64 ]

= - 54.41

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