I would like to compute the derivative of the following parametric equation w.r.t $a$ and $b$:
$P(t,a,b)=a~ \text{cos}(t) i+ b~ \text{sin}(t) j$, with $t \in [0, b]$.
I want to compute the derivative for all values of $t\in [0,b]$
Derivative w.r.t $a$ are easy to compute : $d_a P(t,a,b) = \text{cos}(t)i+ 0j $ with $t \in [0, b]$.
However, the ones w.r.t $b$ are somewhat not intuitive since $t$ depends on $b$.
For example, if $t=0$, $P(0,a,b)=ai+0j $ then $d_b P(0,a,b)=0i+ 0j$
else if $t=b$, $P(b,a,b)=a~ \text{cos}(b)i+ b\text{ sin}(b)j $ then $d_b P(b,a,b)=-a~ \text{sin}(b)i+(\text{sin}(b)+b \text{ cos}(b))j$
I would appreciate if someone has insights on how to compute the derivative w.r.t $b$ for all the values of $t \in [0, b]$.
Thank you.
Reda E.
Given that $t$ is an unknown function of $b$, but I suppose differentiable, I would say: $$ d_bP(t(b),a,b)=-at'\sin(t)i+[\sin(t)+bt'\cos(t)]j $$