Question

Consider F and C below. F(x, y) = x4y5i + x5y4j, C: r(t) = t3 − 2t, t3 + 2t , 0 ≤ t ≤ 1 (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.

Answer 1

$\mathbf f(x,y)=x^4y^5\,\mathbf i+x^5y^4\,\mathbf j$

We want a scalar function $f(x,y)$ whose gradient is equivalent to the vector field. That means

$\dfrac{\partial f}{\partial x}=x^4y^5\implies f(x,y)=\dfrac{x^5y^5}5+g(y)$
$\dfrac{\partial f}{\partial y}=x^5y^4\implies x^5y^4=x^5y^4+\dfrac{\mathrm dg}{\mathrm dy}$
$\implies\dfrac{\mathrm dg}{\mathrm dy}=0\implies g(y)=C$

So (a)

$f(x,y)=\dfrac{x^5y^5}5+C$

which means (b)

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=f(\mathbf r(1))-f(\mathbf r(0))=f(-1,3)-f(0,0)=-\dfrac{243}5-0=-\dfrac{243}5$

Answer 2

(a)$f(x,y)=\frac{x^5y^5}{5} +k$

(b)   $\int\limits_c {f} \, dr=\frac{-243}{5}$

We have $F(x, y) = x^4y^5i + x^5y^4j, C: r(t) = t^3 - 2t, t^3 + 2t , 0 \leq t \leq 1$

(a). As F=∇f

$F(x, y) = x^4y^5i + x^5y^4j$

So,

$\frac{df}{dx} =x^4y^5\\f(x,y)=\frac{x^4y^5}{5} +g(y)\\\frac{df}{dy} =x^5y^4\\f(x,y)=\frac{x^5y^5}{5} +k\\\frac{dg}{dy} =0\\g(y)=k\\f(x,y)=\frac{x^5y^5}{5} +k$

Therefore, $f(x,y)=\frac{x^5y^5}{5} +k$

(b) Now,

$\int\limits_c {f} \, dr=f(r(1)-f(r(0))\\=-f(-1,3)-f(0,0)\\f(x,y)=\frac{x^5y^5}{5}+c \\f(-1,3)=\frac{(-1)^53^5}{5}+c\\f(0,0)=0+c\\\int\limits_c {f} \, dr=\frac{-243}{5} -0\\\int\limits_c {f} \, dr=\frac{-243}{5}$

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