Find dy expressed through dx for the parametric curve x =

Find $dy$ expressed through $dx$ for the parametric curve $x = e^{-0.1t},\ y = \ln t\ \ (t>0)$.

Step 1: Differentiate each coordinate with respect to $t$

\[ x'_{t} = -0.1\,e^{-0.1t}, \qquad y'_{t} = \frac1t. \]

Step 2: Form $\dfrac{dy}{dx}$

\[ \frac{dy}{dx} = \frac{y'_{t}}{x'_{t}} = \frac{1/t}{-0.1\,e^{-0.1t}} = -\frac{10}{t\,e^{-0.1t}} = -\frac{10\,e^{0.1t}}{t}. \]

Step 3: The differential in terms of $t$

\[dy = -\dfrac{10\,e^{0.1t}}{t}\,dx\]

Step 4: The same answer in terms of $x$

The parameter can be eliminated: from $x = e^{-0.1t}$ we get $\ln x = -0.1t$, hence

\[ t = -10\ln x, \qquad e^{0.1t} = \frac1x. \]

Substituting both into the boxed formula, the two factors of $10$ and the two minus signs all cancel: $dy = -\frac{10\cdot\dfrac1x}{-10\ln x}\,dx = \frac{dx}{x\ln x}$.

Step 5: Consistency of the two forms

Eliminating $t$ from the curve itself gives $y = \ln t = \ln(-10\ln x)$. Differentiating this directly: $y' = \frac{1}{-10\ln x}\cdot\left(-\frac{10}{x}\right) = \frac{1}{x\ln x}$.

which matches. Note that $t>0$ forces $0\lt x\lt 1$, so $\ln x\lt 0$ and the slope is negative throughout — consistent with the minus sign in the $t$-form.

Step 6: Verification

\[t = 10:\quad x = e^{-1} \approx 0.367879,\ y = \ln 10 \approx 2.302585\]

\[\frac{dy}{dx} = -\frac{10e}{10} = -e \approx -2.718282; \qquad \frac{1}{x\ln x} = \frac{1}{e^{-1}\cdot(-1)} = -e \;\;\checkmark\]

\[t = 20:\quad x = e^{-2},\ \frac{dy}{dx} = -\frac{10e^{2}}{20} = -\frac{e^{2}}{2} \approx -3.694528 \;\;\checkmark\]