Mathematics & Numerical Analysis

Multivariable chain rule evaluation:
$\frac{d\omega}{dt} = \frac{\partial\omega}{\partial x}\frac{dx}{dt} + \frac{\partial\omega}{\partial y}\frac{dy}{dt}$.
Here $\omega = x^2 y - y^2$:
$\frac{\partial\omega}{\partial x} = 2xy$, $\frac{\partial\omega}{\partial y} = x^2 - 2y$.
Given $x = \sin t \implies \frac{dx}{dt} = \cos t$.
Given $y = e^t \implies \frac{dy}{dt} = e^t$.
At $t = 0$:
$x = \sin 0 = 0$, $y = e^0 = 1$.
$\frac{dx}{dt} = \cos 0 = 1$, $\frac{dy}{dt} = e^0 = 1$.
Substitute these values:
$\frac{\partial\omega}{\partial x} = 2(0)(1) = 0$.
$\frac{\partial\omega}{\partial y} = 0^2 - 2(1) = -2$.
$\frac{d\omega}{dt} = (0)(1) + (-2)(1) = -2$.

Option breakdown:

• $0$ forgets the contribution from the $\partial\omega/\partial y$ term.
• $-1$ is an incorrect calculation.
• $2$ has an incorrect positive sign.
• $-2$ is the correct total derivative evaluated at $t=0$.

Mathematical definition of a cusp versus a vertical tangent:
For a continuous curve $y = f(x)$ at $x = c$:

• Vertical Tangent: Both one-sided limits of the derivative approach the same signed infinity (both $+\infty$ or both $-\infty$). Examples: $f(x) = x^{1/3}$ has $\lim_{x \to 0} f'(x) = +\infty$.
• Cusp: The one-sided limits of the derivative approach opposite signed infinities. That is, as $x \to c^-$, $f'(x) \to +\infty$ and as $x \to c^+$, $f'(x) \to -\infty$ (or vice versa). Example: $f(x) = 1 - x^{2/3}$ or $f(x) = x^{2/3}$.
  
  Option breakdown:
• Both limits tending to $+\infty$ defines an inflection point with a vertical tangent line, not a cusp.
• Both limits tending to $-\infty$ also defines a vertical tangent, not a cusp.
• Opposite infinities ($+\infty$ and $-\infty$) precisely defines a cusp where the curve reverses direction sharply.
• Finite unequal limits define a corner, not a cusp.

Evaluation of limit using L'Hopital's rule:
As $x \to +\infty$, $\ln x \to +\infty$ and $x^2 \to +\infty$, giving an indeterminate form of type $\left[\frac{\infty}{\infty}\right]$.
Applying L'Hopital's rule by differentiating numerator and denominator with respect to $x$:
$\lim_{x \to +\infty} \frac{\frac{d}{dx}(\ln x)}{\frac{d}{dx}(x^2)} = \lim_{x \to +\infty} \frac{1/x}{2x} = \lim_{x \to +\infty} \frac{1}{2x^2}$.
As $x \to +\infty$, the denominator $2x^2 \to +\infty$, so:
$\frac{1}{2x^2} \to 0$.

Option breakdown:

• $1$ is incorrect.
• $-\infty$ is incorrect.
• $+\infty$ would occur if the power of $x$ were inverted.
• $0$ is the exact limit.