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13.1 - Potentials - finding the gradient fields
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<h1>Finding the potentials</h1>
<p>The following vector fields are all conservative.  For each one, find a potential function, $f(x,y)$ or $f(x,y,z)$, such that $\myv \grad f=\myv F$.
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<li id="1">$\myv F(x,y)=3xy^2\uv i + 3x^2y\uv j$<br>
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<p id="a1" class="answer toggle">
$$f(x,y)=\frac 32x^2y^2$$</p>
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<li id="2">$\myv F(x,y)=y\sin(xy)\uv i + x\sin(xy)\uv j$<br>
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$$f(x,y)=-cos(xy)$$</p>
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<li id="3">$\myv F(x,y)=(2x+y)\uv i + (x+3y^2)\uv j$<br>
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<p id="a3" class="result toggle">
Let's see... try integrating $P=2x+y$:  $\int (2x+y)\,dx=x^2+xy$.  Maybe $f(x,y)=x^2+xy$?<br><br>
Taking the partial derivative w.r.t. $y$:
$f_y=x$.  But we have $Q=x+y^2$.  The $x$ term is accounted for.  So, try adding something like $\int y^2=y^3/3$?  That is, how about
$$g(x,y)=x^2+xy+\frac {y^3}{3}?$$
Now $g_y=x+y^2$: Yes! that's $Q$.  
Check it for $P$:
$$g_x=2x+y$$
Yes, that's $P$! So the answer is <span class="answer">$g(x,y)=x^2+xy+\frac{y^3}{3}$</span>.</p>


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<li id="4">$\myv F(x,y,z)=yze^{xyz}\uv i + xze^{xyz}\uv j + xye^{xyz}\uv k$<br><a href="#4" onclick="toggle_visibility('a4');">show / hide</a>
<p id="a4" class="answer toggle">
$$f(x,y,z)=e^{xyz}$$
<span class="result">E.g. $\frac{\del}{\del x}e^{xyz}=e^{xyz}\frac{\del}{\del x}(xyz)=e^{xyz}yz$.
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