Find the slope of the tangent to the curve x = t ln t, y

Find the slope of the tangent to the curve $x = t\ln t$, $y = \dfrac{\ln t}{t}$ at the point $(0,0)$.

Step 1: Which value of $t$ gives the origin?

The parametrisation requires $t \gt 0$. Two candidates suggest themselves, and only one works.

The limit $t \to 0^{+}$ gives $x = t\ln t \to 0$, which looks promising — but the second coordinate runs away:

\[ \lim_{t\to 0^{+}}\frac{\ln t}{t} = \frac{-\infty}{0^{+}} = -\infty. \]

So that end of the curve descends to $-\infty$ along the $y$-axis; it does not reach the origin.

The value $t = 1$ makes both coordinates vanish at once, since $\ln 1 = 0$:

\[ x(1) = 1\cdot\ln 1 = 0, \qquad y(1) = \frac{\ln 1}{1} = 0. \;\checkmark \]

So the origin corresponds to $t = 1$.

Step 2: Differentiate with respect to $t$

Product rule for $x$, quotient rule for $y$: \[x'_{t} = \ln t + t\cdot\frac1t = \ln t + 1, \qquad y'_{t} = \frac{\frac1t\cdot t - \ln t}{t^{2}} = \frac{1-\ln t}{t^{2}}.\]

Step 3: Form the ratio

\[ \frac{dy}{dx} = \frac{1-\ln t}{t^{2}\bigl(\ln t + 1\bigr)}. \]

Step 4: Evaluate at $t = 1$

With $\ln 1 = 0$: $k = \frac{1-0}{1\cdot(0+1)} = 1$.

Step 5: Verification

\[t = 1.001:\quad x \approx 0.0010005,\ y \approx 0.0009985 \;\Longrightarrow\; \frac{\Delta y}{\Delta x} \approx 0.998\]

\[t = 0.999:\quad x \approx -0.0009995,\ y \approx -0.0010015 \;\Longrightarrow\; \text{slope} \approx 1.002 \;\;\checkmark\]

The tangent at the origin is the line $y = x$.