It's a matter of degree--the sawtooth decline on a low-ripple PSU is usually pretty straight simply because you are only seeing the very beginning of the decline. If you turned the power off and watched the voltage ebb away, it would indeed have a curve.
Sure, but that's highly dependent upon the power supply, no? In particular, the RC time constant versus the ripple frequency.
Here, the ripple frequency is only 100 Hz, for a 10 ms period. That seems to me to be a pretty long period for ripple, no? The peak to peak ripple is about 120 mV. This translates to a slew rate of 24V/sec, which seems rather slow.
In light of that, my sense is that the curvature seen is unsurprising.
So it is pretty obviously a 50Hz full-bridge cap-filtered supply. Let's assume the load is resistive. The ripple appears to be 134mV, but the OP hasn't given the other details, so let's assume that this is a 13.4V supply, and thus the ripple represents 1% of the total voltage. This means that the peak current will be 13.4V/R
LOAD and the minimum current will be 0.99 x that. OK?
So the dV/dt will be proportional to the current during the capacitor discharge (the downslope) and is represented by the angle of the tangential to the trace at any one point, since the vertical difference is the dV and the horizontal dt. The actual angle will depend on the selection of V/div and s/div, but let's say we adjust those so that whatever the peak dV/dt is, it is represented by a 45 degree angle on the scope. The tangent of this angle is 1, representing dV/dt * some constant, and thus the tangent of the angle at the bottom of the slope, dV/dt
min will be 0.99*dV/dt
peak. Arctan (0.99) = ~44.7 degrees, so a 0.3 degree difference in the slope from the peak to the bottom is what you'd expect. That's pretty straight. I hope I didn't screw up the math and that it wasn't too unclear.