Why the v-t and i-t curves for capacitor are exponential curve?
Capacitance is defined as:

\$C = \frac{Q}{V}\$

and equally:

\$C = \frac{\delta Q}{\delta V}\$

Since current is defined as the rate of change of charge:

\$I(t) = \frac{\delta Q(t)}{\delta t}\$

Thus:

\$I(t) = C \frac{\delta V(t)}{\delta t}\$

If we set up a circuit with a voltage source, Resistor and a capacitor

\$V = V_r + V_c\$ and this creates a differential equation

\$v = i(t)R + \frac{1}{C}\int i(t) \delta t  \$

Solving such differential equations produces an equation:

\$I(t) = \frac{V}{R}e^\frac{-t}{RC}\$

Which equally can be re-arranged to be

\$V(t) = V(1-e^\frac{-t}{RC}) \$ taking into account initial conditions