# Derivative parameric equation

I would like to compute the derivative of the following parametric equations w.r.t $$a$$ and $$b$$:

$$x=a~ \text{cos}(t)$$ and $$y= b~ \text{sin}(t)$$ with $$t \in [0, b]$$.

Derivative w.r.t $$a$$ are easy to compute : $$d_a x = \text{cos}(t)$$, $$d_a y = 0$$ with $$t \in [0, b]$$.

However, the ones w.r.t $$b$$ are somewhat not intuitive since $$t$$ depends on $$b$$.

For example, if $$t=b$$ then $$d_b x =-ab \text{sin}(b)$$

I would appreciate if someone has insight on how to compute these derivatives.

Thank you.

Reda E.

• $t$ doesn't depend on $b$; it's still an independent variable. Its range depends on $b$ but that's not the same thing. Commented Apr 17, 2020 at 18:59
• Actually it does, because $d_b x = \partial_b x + \partial_t x d_b t$. The last term $d_b t$ shows the dependence.
– Reda
Commented Apr 17, 2020 at 19:34
• You must first clarify which are the dependent and independent variables.
– amd
Commented Apr 17, 2020 at 22:54

On the contrary, $$t$$ doesn't depend on $$b.$$ Or at least not for most of its values. It depends on $$b$$ only once in the interval $$[0,1/b].$$ Otherwise it doesn't. So the derivatives are as before.
• Actually it does, because $d_b x = \partial_b x + \partial_t x d_b t$. The last term $d_b t$ shows the dependence.
• @Reda I see you've changed that interval. In any case, as I've said, $t$ is a function of $b$ only at the right endpoint of your interval. If you need the derivative at this endpoint, set $t=b$ and differentiate. Commented Apr 18, 2020 at 5:58
• what if $t=b-\varepsilon$, it is also a special case