
Hi! This was my homework. Prof. sent its answer. But I didnt understand how can this answer be reached? Please can someone explain this?

Hi! This was my homework. Prof. sent its answer. But I didnt understand how can this answer be reached? Please can someone explain this?
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Note that $i(u,v) = (u,v,\sqrt{1-u^2-v^2})$. Now we calculate:
$$i_*\left(\left.\frac{\partial}{\partial u}\right|_p\right)(a) = \frac{\partial}{\partial u}(a\circ i) = \frac{\partial}{\partial u}(u) = 1$$ $$i_*\left(\left.\frac{\partial}{\partial u}\right|_p\right)(b) = \frac{\partial}{\partial u}(b\circ i) = \frac{\partial}{\partial u}(v) = 0$$ $$i_*\left(\left.\frac{\partial}{\partial u}\right|_p\right)(c) = \frac{\partial}{\partial u}(c\circ i) = \frac{\partial}{\partial u}(\sqrt{1-u^2-v^2}) = \frac{-u}{\sqrt{1-u^2-v^2}} = -\frac{a}{c}$$
The pushforward of $\frac{\partial}{\partial v}$ is similar.