Find \(dy/dx\) where \(y=f(x)\) is given by the parametric equations \(x=t+5\cos(t)\), \(y=3e^t\) .

Find the derivative of the parametric curve \(x=t\sin(t)\), \(y=t^2+t\).

Find \(dy/dx\) of the parametric curve \(x=4t+1\), \(y=t^2+2t\).

Find the second derivative \(\dfrac{d^2y}{dx^2}\) where \(y=f(x)\) is given by the parametric equations \(x=t^2+t\), \(y=2t-1\) .

Find the second derivative of the parametric curve \(x=t^3+t\), \(y=t^5+1\).

Find the second derivative of the parametric curve \( x = t - t^3 \), \( y = 2t + 5 \).

Find the horizontal tangent points to the curve \(x=1-t, y=t^2\) and determine the concavity at those points.

How do you know when you have a horizontal tangent? When the slope is zero.
\(dx/dt=-1, dy/dt=2t\)
\(\displaystyle{ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t}{-1} }\)
\(dy/dx = 0\) when \(-2t=0 \to t=0\)
When \(t=0\), \((x,y) = (1,0)\).

To determine concavity at that point, we need to calculate the second derivative.
\(\displaystyle{ \frac{d^2y}{dx^2} = \frac{d[dy/dx]/dt}{dx/dt} = \frac{-2}{-1} = 2 }\)
Since the second derivative is positive everywhere, the curve is concave up.

\((1,0)\), concave up

For the parametric curve \(x=\sqrt{t}, y=3t-1\), find the derivative and the slope at the point \(t=1\). Also determine the concavity and the equation of the tangent line at that same point. Give your equation of the tangent line in slope-intercept form.

\(dx/dt=(1/2)t^{-1/2}, dy/dt = 3\)
\(\displaystyle{ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3}{(1/2)t^{-1/2}} = 6\sqrt{t} }\)
slope at \(t=1, m=6\sqrt{1} = 6\)
for the equation of tangent line, we need the \((x,y)\) point at \(t=1\), so \((\sqrt{1},3-1)=(1,2)\)
\(y-y_1=m(x-x_1) \to y-2=6(x-1) \to y=6x-4\)
\(\displaystyle{ \frac{d^2y}{dx^2} = \frac{d[dy/dx]/dt}{dx/dt} = \frac{6(1/2)t^{-1/2}}{(1/2)t^{-1/2}} = 6 }\)
Since the second derivative is positive at \(t=1\) (and everywhere else for that matter), the curve is concave up.

\(dy/dx=6\sqrt{t}, m=6, y=6x-4\), concave up