Introduces fundamental calculus concepts for parametric equations, focusing on finding the slope of a tangent line (dy/dx) and identifying horizontal and vertical tangents.
Given parametric equations \(x = f(t)\) and \(y = g(t)\), how do you find the derivative \(\frac{dy}{dx}\) without eliminating the parameter?
You use the chain rule. The formula is \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\), provided that \(dx/dt \neq 0\).
Find \(\frac{dy}{dx}\) for the curve given by \(x = t^2\) and \(y = t^3\).
First, find the derivatives with respect to \(t\): \(\frac{dx}{dt} = 2t\) and \(\frac{dy}{dt} = 3t^2\).
Then, \(\frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3}{2}t\), for \(t \neq 0\).
How do you find the points where a parametric curve has a horizontal tangent line?
A horizontal tangent occurs when the slope \(\frac{dy}{dx}\) is \(0\). This happens when \(\frac{dy}{dt} = 0\), as long as \(\frac{dx}{dt} \neq 0\) at the same value of \(t\).
How do you find the points where a parametric curve has a vertical tangent line?
A vertical tangent occurs when the slope \(\frac{dy}{dx}\) is undefined. This happens when \(\frac{dx}{dt} = 0\), as long as \(\frac{dy}{dt} \neq 0\) at the same value of \(t\).
Find the slope of the tangent line to the curve \(x = 4\cos(t)\), \(y = 3\sin(t)\) at \(t = \frac{\pi}{4}\).
\(\frac{dx}{dt} = -4\sin(t)\) and \(\frac{dy}{dt} = 3\cos(t)\). So, \(\frac{dy}{dx} = \frac{3\cos(t)}{-4\sin(t)} = -\frac{3}{4}\cot(t)\).
At \(t = \frac{\pi}{4}\), the slope is \(-\frac{3}{4}\cot(\frac{\pi}{4}) = -\frac{3}{4}(1) = -\frac{3}{4}\).
Find the values of \(t\) for which the curve \(x = t^3 - 3t\), \(y = t^2 - 4\) has a horizontal tangent.
Set \(\frac{dy}{dt} = 0\). \(\frac{dy}{dt} = 2t = 0\), which gives \(t=0\). At \(t=0\), \(\frac{dx}{dt} = 3t^2-3 = -3 \neq 0\). So, a horizontal tangent exists at \(t=0\).
Find the values of \(t\) for which the curve \(x = t^3 - 3t\), \(y = t^2 - 4\) has a vertical tangent.
Set \(\frac{dx}{dt} = 0\). \(\frac{dx}{dt} = 3t^2 - 3 = 0\), so \(3(t^2-1)=0\), which gives \(t=1\) and \(t=-1\). At these values, \(\frac{dy}{dt} = 2t \neq 0\). So, vertical tangents exist at \(t=1\) and \(t=-1\).
What is the formula for the second derivative, \(\frac{d^2y}{dx^2}\), of a parametric curve?
The formula is \(\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}\). You must first find \(\frac{dy}{dx}\), then differentiate that result with respect to \(t\), and finally divide by \(\frac{dx}{dt}\).
Why is \(\frac{d^2y}{dx^2}\) NOT equal to \(\frac{d^2y/dt^2}{d^2x/dt^2}\)?
Because \(\frac{d^2y}{dx^2}\) means \(\frac{d}{dx}(\frac{dy}{dx})\). Applying the chain rule to the derivative operation itself leads to the correct, more complex formula: \(\frac{d}{dt}(\frac{dy}{dx}) \cdot \frac{dt}{dx}\).
Find \(\frac{d^2y}{dx^2}\) for the curve given by \(x = t^2\) and \(y = t^3\).
Find the point(s) where the curve given by \(x = 1+\sin(t)\), \(y = \cos(t)\) for \(0 \le t < 2\pi\) has a horizontal tangent.
\(y'(t) = -\sin(t)\). Setting \(-\sin(t) = 0\) gives \(t=0\) and \(t=\pi\). At these times, \(x'(t) = \cos(t) \neq 0\). The points are \((1, 1)\) at \(t=0\) and \((1, -1)\) at \(t=\pi\).
Find the equation of the tangent line for \(x=2t, y=t^2+5\) at \(t=1\).
At \(t=1\), the point is \((2, 6)\). The slope is \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t}{2} = t\). At \(t=1\), the slope is \(1\). Using point-slope form: \(y - 6 = 1(x - 2)\), so \(y = x + 4\).
What does a point where \(\frac{dy}{dt} = 0\) and \(\frac{dx}{dt} = 0\) signify on a parametric curve?
This indicates a point where the slope is indeterminate (\(\frac{0}{0}\)). Such points are often cusps or sharp corners on the curve, and the tangent line's existence must be investigated further, often by taking a limit.
