Parametric Equations & Derivatives - Calculus 2

1. Consider the parametric curve $x = a\cos^3(t)$ and $y = a\sin^3(t)$ where $a > 0$. If the length of the curve from $t = 0$ to $t = \frac{\pi}{2}$ is $\frac{3a}{2}$, what is the slope of the curve at $t = \frac{\pi}{4}$?
   Answer: $-1$ First, find the derivatives: $\frac{dx}{dt} = -3a\cos^2(t)\sin(t)$ and $\frac{dy}{dt} = 3a\sin^2(t)\cos(t)$. The slope is $\frac{dy}{dx} = \frac{3a\sin^2(t)\cos(t)}{-3a\cos^2(t)\sin(t)} = \frac{\sin(t)}{-\cos(t)} = -\tan(t)$. At $t = \frac{\pi}{4}$: $\frac{dy}{dx} = -\tan\left(\frac{\pi}{4}\right) = -1$. The arc length condition is given but doesn't affect the slope calculation - it's extra information that might confuse students. The arc length from $t = 0$ to $t = \frac{\pi}{2}$ would be $\int_0^{\pi/2} \sqrt{(dx/dt)^2 + (dy/dt)^2} dt = \int_0^{\pi/2} 3a\sin(t)\cos(t) dt = \frac{3a}{2}$, which confirms the given information but is irrelevant to finding the slope. Students might incorrectly try to use this information, leading to wrong answers like B or D.
2. For the parametric equations $x = \ln(1 + t^2)$ and $y = \arctan(t)$, what is the value of $\frac{dy}{dx}$ when $x = \ln(2)$?
   Answer: $\frac{1}{2}$ When $x = \ln(2)$, we have $\ln(1 + t^2) = \ln(2)$, so $1 + t^2 = 2$, giving $t^2 = 1$, thus $t = \pm 1$. We have $\frac{dx}{dt} = \frac{2t}{1 + t^2}$ and $\frac{dy}{dt} = \frac{1}{1 + t^2}$. Therefore, $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{1}{1 + t^2}}{\frac{2t}{1 + t^2}} = \frac{1}{2t}$. For $t = 1$: $\frac{dy}{dx} = \frac{1}{2(1)} = \frac{1}{2}$. For $t = -1$: $\frac{dy}{dx} = \frac{1}{2(-1)} = -\frac{1}{2}$. Since the question asks for 'the value' without specifying which branch, and choice A is $\frac{1}{2}$, this corresponds to $t = 1$. The other choices represent common errors: B is $\frac{1}{4}$ (forgot the factor of 2), C involves $\sqrt{2}$ (confusion with the value $t^2 = 1$), and D combines these errors.
3. A parametric curve is defined by $x = t^4 - 4t^2$ and $y = t^3 - 3t$. At which value(s) of $t$ does the curve have the steepest slope (greatest absolute value of $\frac{dy}{dx}$)?
   Answer: $t = \pm \sqrt{\frac{3}{2}}$ We have $\frac{dx}{dt} = 4t^3 - 8t = 4t(t^2 - 2)$ and $\frac{dy}{dt} = 3t^2 - 3 = 3(t^2 - 1)$. So $\frac{dy}{dx} = \frac{3(t^2 - 1)}{4t(t^2 - 2)}$. To find the maximum of $\left|\frac{dy}{dx}\right|$, we need to find critical points of $f(t) = \frac{3(t^2 - 1)}{4t(t^2 - 2)}$. Using the quotient rule and setting $f'(t) = 0$ leads to a complex calculation. Alternatively, we can analyze $\left(\frac{dy}{dx}\right)^2 = \frac{9(t^2 - 1)^2}{16t^2(t^2 - 2)^2}$ and find its maximum. Taking the derivative and setting to zero: $\frac{d}{dt}\left[\frac{(t^2 - 1)^2}{t^2(t^2 - 2)^2}\right] = 0$. This gives $t^2 = \frac{3}{2}$, so $t = \pm\sqrt{\frac{3}{2}}$. Choice A gives $t^2 = \frac{2}{3}$, choice B gives $t^2 = 1$, and choice D gives $t^2 = 3$, which are incorrect critical points.
4. A curve is defined parametrically by $x = t^3 - 3t$ and $y = 2t^2 - 1$. At which value of $t$ does the curve have a horizontal tangent line?
   Answer: $t = 0$ only For a horizontal tangent, we need $\frac{dy}{dx} = 0$. Using $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$, we have $\frac{dy}{dt} = 4t$ and $\frac{dx}{dt} = 3t^2 - 3$. For horizontal tangent: $\frac{dy}{dt} = 0$ and $\frac{dx}{dt} \neq 0$. From $4t = 0$, we get $t = 0$. At $t = 0$: $\frac{dx}{dt} = 3(0)^2 - 3 = -3 \neq 0$. So $t = 0$ gives a horizontal tangent. Choice B gives vertical tangents (where $\frac{dx}{dt} = 0$). Choices C and D incorrectly include additional values.

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