adjusting swept signal equation
The general formula for the exponential rise in frequency is given by

$f=A+B(1-\exp(-\alpha t))$

We need to solve for the values of $A$, $B$ and $\alpha$. Let's assume you want an initial frequency of $20$ at time $t=0$, and an asymptotic frequency of $200$ Hz at time $t=\infty$.

Substituting $t=0$ and $f=20$ into the above equation we find $A=20$. If we substitute $t=\infty$ and $f=200$ we find that $B=180$. For $\alpha$ you really have a free choice - it controls how fast the frequency rises. If you have another control point say at time $t=2$ we want the frequency to be at $95%$ of the final frequency i.e. $f=.95 \times 200=190$ then we can substitute these into the above equation (Since we know what $A$, and $B$ are).

$190=20+180(1-\exp(-2\alpha))$

Solving gives $\alpha= -0.5\ln(10/180)=-1.445$. If you want a faster rise then you'll need to choose a larger value of $\alpha$.

To find the expression for phase we simply integrate with respect to time to give

$\phi =2\pi(At+Bt+\frac{B}{\alpha}\exp(-\alpha t))$