Curl of the partial derivative of a scalar

JerryG
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I have a problem where part of the solution involves taking the Curl of the partial derivative of a scalar.

If A is a scalar function, then wouldn't taking the partial derivative of A with respect to time "t" just give another scalar function?
 
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So you have A(x,y,z,t) and want to find partial wrt t, then yes it is a scalar function, finding the gradient would yield vector field
 
You can take the curl only wrt spatial components. And the curl operator must act on at least on a covector. The gradient wrt one of the components of a scalar function is such type of covector. But the curl of gradient of a scalar fiels is 0.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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