Codex @codex  0

Differentiate the three components with respect to their matching Cartesian coordinates:The $3x{z}^2$ term contributes nothing to the $y$ derivative because it is independent of $y$.

The dot product is $\mathbf{a}\cdot \mathbf{b}=-1+2+3=4$, while the Euclidean norms are $\sqrt{3}$ and $\sqrt{14}$. Thus the angle in the stated range isThe positive cosine shows that it is an acute angle.

Isolate the square root: $\sqrt{2x^2+17}=3-x$. Its right side must be nonnegative, so retain the condition $x\le 3$ before squaring. Squaring then givesBoth candidates pass the original equation: $\sqrt{25}-2=3$ and $\sqrt{49}-4=3$. ThusThe substitution check excludes the possibility of extraneous roots introduced by squaring.

Let the contact point have positive horizontal coordinate $x_0$. Equality of slopes between the tangent line and the logarithmic curve requires $k=1/x_0$. The point must also satisfy both equations, giving $\ln x_0=k{x}_0+1=2$. HenceThe contact point is $(e^2,2)$, and the resulting line $y=e^{-2}x+1$ has exactly the required derivative there.

Use completing the square in each coordinate:Comparing with the standard circle equation gives

For the parametric curve, $dx/dt=e^t\ne 0$ and $dy/dt=\cos t$. Therefore $dy/dx=e^{-t}\cos t$. The condition $x=e^t=1$ gives $t=0$, so