Answer
Equation of the tangent line: $y = \frac 1 6 x$
Graph: The purple curve is the tangent line, and the green one is the parametric equation:
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Work Step by Step
1. Calculate $\frac{dy}{dt}$ and $\frac{dx}{dt}$:
$\frac{d(t^2+t)}{dt} = 2t + 1$
$\frac{d(6sin(t))}{dt} = 6cos(t)$
Therefore: $\frac{dy}{dx} = \frac{2t + 1}{6cos(t)}$
2. At which point $x = 0$ and $y = 0$?
$0 = 6cos(t)$
$0 = t^2 + t$
- The only answer that satisfies both equations is $t = 0$
3. Calculate $\frac{dy}{dx}$ for t = 0:
$\frac{2(0) + 1}{6cos(0)} = \frac 1 {6*(1)} = \frac 1 6$
4. Determine the tangent equation:
$y-y_0 = \frac{dy}{dx}(x-x_0)$
$y - 0 = \frac 1 6 (x - 0)$
$y = \frac 1 6 x$
