Answer
Estimated coordinates:
Leftmost point: $(-1.2, 1.2)$
Lowest point: $(1.5, -0.5)$
Exact coordinates:
Leftmost point: $(-1.19, 1.19)$
Lowest point: $(1.42, -0.47)$
Work Step by Step
1. Graph the equation and estimate the coordinates of the leftmost and the lowest point on that curve.
- Insert the parametric equation and then zoom in the leftmost and in the lowest point, and take note of the coordinates.
- Mine were $(-1.2,1.2)$ and $(1.5,-0.5)$, respectively.
2. Calculate $\frac{dy}{dt}$ and $\frac{dx}{dt}$:
$\frac{dy}{dt} = \frac{d(t + t^4)}{dt} = 1 + 4t^3$
$\frac{dx}{dt} = \frac{d(t^4-2t)}{dt} = 4t^3 - 2$
Thus: $\frac{dy}{dx} = \frac{1 + 4t^3}{4t^3 - 2}$
The leftmost point is at a vertical tangent, therefore, $4t^3 - 2= 0$ and $1 + 4t^3 \neq 0$
$4t^3 - 2= 0$
$4t^3 = 2$
$t^3 = \frac 2 4 = \frac 1 2$
$t = \sqrt [3] {\frac 1 2}$
- Check if $1 + 4t^3 \neq 0$:
$1 + 4(\sqrt [3] {\frac 1 2})^3 \neq 0 $
$1 + 4(\frac 1 2) \neq 0$
$3 \neq 0$ - It is valid.
The lowest point is at a horizontal tangent, therefore, $1 + 4t^3 = 0$ and $4t^3 -2 \neq 0$
$1 + 4t^3 = 0$
$4t^3 = -1$
$t^3 = - \frac{1}{4}$
$t =- \sqrt [3] {\frac 1 4}$
- Check if: $4t^3 - 2 \neq 0 $
$4(\sqrt [3] {\frac 1 4})^3 -2 \neq 0$
$1 - 2 \neq 0 $
$-1 \neq 0$ - It is also valid.
3. Find the coordinates for: $t = \sqrt [3] {\frac 1 2}$
$x = (\sqrt [3] {\frac 1 2})^4 - 2(\sqrt [3] {\frac 1 2}) = -1.19$
$y = \sqrt [3] {\frac 1 2} + (\sqrt [3] {\frac 1 2})^4 = 1.19$
4. Find the coordinates for: $t = \sqrt [3] {\frac 1 4}$
$x = (-\sqrt [3] {\frac 1 4})^4 - 2(-\sqrt [3] {\frac 1 4}) = 1.42$
$y = -\sqrt [3] {\frac 1 4} + (-\sqrt [3] {\frac 1 4})^4 = -0.47$
