Find the slope of the tangent to the curve x = t³+1, y =

Find the slope of the tangent to the curve $x = t^{3}+1$, $y = t^{2}+t+1$ at the point $(1,1)$.

Step 1: Find the parameter value

From $x = t^{3}+1 = 1$ we get $t^{3} = 0$, so $t = 0$ — the only real solution. Checking: $y = 0+0+1 = 1$. $\checkmark$

Step 2: Differentiate with respect to $t$

\[ x'_{t} = 3t^{2}, \qquad y'_{t} = 2t+1. \]

At $t = 0$ the first vanishes while the second does not: $x'_{t}(0) = 0$ and $y'_{t}(0) = 1$. This is the signature of a vertical tangent.

Step 3: Form the ratio and take the limit

\[ \frac{dy}{dx} = \frac{2t+1}{3t^{2}}. \]

As $t \to 0$ the numerator tends to $1$ while the denominator tends to $0$ through positive values (since $3t^{2} \gt 0$ on both sides), so

\[ \lim_{t\to 0}\frac{2t+1}{3t^{2}} = +\infty \]

from the left as well as from the right.

Step 4: Verification

\[t = 0.01:\quad \frac{dy}{dx} = \frac{1.02}{0.0003} = 3400\]

\[t = -0.01:\quad \frac{dy}{dx} = \frac{0.98}{0.0003} \approx 3267 \text{ — also large and positive} \;\;\checkmark\]

Step 5: What the curve looks like near the point

Solving $t = \sqrt[3]{x-1}$ and substituting gives

\[ y = (x-1)^{2/3} + (x-1)^{1/3} + 1, \]

and the term $(x-1)^{1/3}$ is what produces the vertical tangent — the same cube-root behaviour as in Problem 24b.