1
$\begingroup$

Let $f: \mathbb{R^2} \to \mathbb{R^2}$ be a differentiable function in $(0,0)$ and $\gamma_1, \gamma_2 :[a,b] \to \mathbb{R^2}$ differentiable paths such that $\gamma_1(0)=\gamma_2(0)=(0,0)$ and we know $\gamma_2 (t)=\gamma _1(t)+(t^2,t^3) $

I need to show that $(i)$ $\frac{d(f(\gamma_1 (t)))}{dt}|_{t=0}=\frac{d(f(\gamma_2 (t)))}{dt}|_{t=0}.$

I defined $g1=f((\gamma_1(t))) $ and $ g2=f((\gamma_2(t)))$, If I'm not mistaken I need to show that $\frac{d g_1}{dt}(0)=\frac{dg_2}{dt}(0)$, and If I'm not mistaken $(ii)$ $\frac{d g_1}{dt}=\frac{df}{dx}(\gamma_1(0))\frac{dx_1}{dt}+\frac{df}{dy}(\gamma_1(0))\frac{dy_1}{dt},$ $\frac{d g_2}{dt}=\frac{df}{dx}(\gamma_2(0))\frac{dx_2}{dt}+\frac{df}{dy}(\gamma_2(0))\frac{dy_2}{dt},$ (Are these right definitons?)

where $\gamma_1(t)=(x_1(t),y_1(t))$, $\gamma_2(t)=(x_1(t),y_2(t))$ (Did I get right these as well?)

Now we know as well $\frac{dx_2 }{dt}=\frac{dx_1}{dt}+t, \frac{dy_2 }{dt}=\frac{dy_1}{dt}+3t^2,$

Can I just plug in $\frac{dx_2 }{dt}(0)=\frac{dx_1}{dt}(0)+0, \frac{dy_2 }{dt}(0)=\frac{dy_1}{dt}(0)+30^2,$ and then use it in and conclude that $(ii)$ is correct and due to that $(i)$ is correct also?

What do I do with all this information? I think I got it somehow wrong. I'd really love your guidance.

Thanks a lot!

1 Answers 1

1

Why do you think it's wrong? If you show one set of quantities are equal to another set, then you may legimitately interchange the two sets of quantities in any equation. In particular, if you write out $df\circ\gamma_2/dt|_{t=0}$ with the chain rule, substitute certain parts of the written-out expression using the fact that \gamma_1'(0)=\gamma_2'(0) and then note the resulting expression is $df\circ\gamma_1/dt|_{t=0}$ evaluated also with the chain rule, then you have shown that these two are in fact equal.