calculusgraphing-functions
My teacher assigned us some graphing homework. I know how to graph all of them manually all except these two
$f_1(x)=x^4-x^3-4x^2$
$f_2(x)=\frac{1}{2}x^3+2x^2-8x-2$
When i asked him how he said it was above our grade level and it was a calculus question so we should just use graphing calculators. So since I'm still curious, I'm asking here. Detailed answers with steps would be nice. The main thing I need to know is how to find the exact location of turning points.
Upper figure: cylinder oil tank cross-section perpendicular to its horizontal axe. The vertical coordinate is the oil level in percentage.
Lower figure: graph of oil volume/max. volume (in %) versus oil level $l$ (in feet). The horizontal straight lines represent the area/volume ratio $A(l)/A(10)=V(l)/V(10)$ (in %) for every multiple of 10% from 0% to 100%.
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Since the tank radius is $5$, the oil level with respect to the bottom of the tank is given by $l=5-5\cos \frac{\theta }{2}$, where $\theta $ is the central angle as shown in the figure. The area of the tank cross section filled with oil is
$$A(\theta )=\frac{25}{2}\theta -\frac{25}{2}\sin \theta $$
or
$$A(l)=25\arccos (\frac{5-l}{5})-\frac{25}{2}\sin (2\arccos (\frac{5-l}{5}))$$
The area ratio $A(l)/A(10)=V(l)/V(10)$ where $V(l)$ is the oil volume.
Let $f(l)$ denote this area ratio in percentage:
$$f(l)=\frac{100}{\pi }\arccos \left( 1-\frac{1}{5}l\right) -\frac{50}{\pi }\sin \left( 2\arccos \left( 1-\frac{1}{5}l\right) \right) $$
Here is the sequence of $f(l)$ values for $l=0,1,2,\ldots ,10$. The graph of $f(l)$ is shown above.
$f(0)=0$, $f(1)=5.2044$, $f(2)=14.238$, $f(3)=25.232$, $f(4)=37.353$, $f(5)=50$,
$f(6)=62.647$, $f(7)=74.768$, $f(8)=85.762$, $f(9)=94.796$, $f(10)=100$
Edit: One still needs to solve the nonlinear equation $f(l)-10k=0$ for $k=1,2,3,4,6,7,8,9$, e. g. by the Secant Method.
Edit 2: The problem of solving graphically, as shown in the secong figure, is that it would be very difficult, or impossible, to get the required accuracy of 0.01 (feet).
Update: The oil level marks (in feet) should be placed at
$0,1.57,2.54,3.40,4.21,$
$5,5.79,6.60,7.46,8.44,10$
corresponding to the oil volume percentage of
$0,10,20,30,40,$
$50,60,70,80,90,100$.
This calculation was based on the following $f$ function values:
$f(0)=0.0$, $f(1.5648)=10.0$, $f(2.5407)=20.0$, $f(3.40155)=30.0$, $f(4.21135)=40.0$
$f(5)=50.0$, $f(5.7887)=60.0$, $f(6.59845)=70.000$, $f(7.4593)=80.000$,
$f(8.4352)=90.000$, $f(10)=100.0$
Update 2 Figure of marks:
[Rearranged to show the sequence of editions and updates.]
Consider the simpler problem of two parametrized curves $(x_i(t),y_i(t))$ that start at $(0,0)$ at time $t=0$, and equality of trajectories up to reparametrization, which is stronger than equality of point sets and a more natural condition as it is local (up to matching of starting points). Heuristically, and to some extent rigorously, there is a usable criterion.
We want that when $x_1 = x_2$, then $y_1 = y_2$ so that the bivariate function $x_1(t)-x_2(s)$ divides $y_1(t)-y_2(s)$ (and vice versa) in a suitable ring of functions. Their ratio is an invertible function with positive values, at least for nonzero $s,t$ near $0$. In fact we need it to be positive only for nonzero $(s,t)$ at which $x_1(s)=x_2(t)$ or the same for $y$.
Example: parabola and half-parabola.
Curve A is $(t,t^2)$.
Curve B is $(s^2,s^4)$.
$x_1(t)-x_2(s) = t - s^2$
$y_1(t)-y_2(s) = t^2 - s^4$
Ratio is $(t + s^2)$
This is positive near the locus where $x_1(t)=x_2(s)$ (namely $t=s^2$). On the locus where $y_1(t)=y_2(s)$ (namely $t^2=s^4$), this is positive for $t>0$ and negative for $t < 0$. The positivity condition knows which half of the parabola is curve B! That is a good sign that this is either the complete answer to the simplified problem, or on the right track.
Finding an intersection point of two parametric curves OR detecting a difference between the curves is simpler than the general problem of curve intersection. Take a point on one curve, solve for the parameter values that would place its $x$ coordinate on the other curve, and test whether the $y$ coordinates are the same. For algebraic parameterizations this calculation can be done exactly.
For the point-set equality problem, locate the zeros of the $st$ ratio. These parameter values segment the two curves into arcs. Then there is a combinatorial problem of orienting and matching (by the procedure given above) identical arcs of the two curves, and testing whether both curves are covered by the matched arcs.
Best Answer
Let's just take $f_1(x) = x^4-x^3-4x^2 = x^2\ (x^2 - x - 4)$. Start by finding the roots (the values of $x$ where $f_1(x) = 0$). A quartic equation may have up to four real roots. From the $x^2$ factor, we see that two of them are $x = 0$ and $x = 0$ — a "double" root.
What about the other two roots? To solve $x^2 - x - 4 = 0$, we can use the quadratic equation:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm \sqrt{17}}{2} $$
So, $y = x^2 - x - 4$ would describe a parabola that crosses the $x$-axis at -1.562… and 2.562…. It is centered at $x = \frac{1}{2}$ (the midpoint between the roots), with a minimum at (0.5, -4.25).
Now let's consider what happens when we multiply that by $x^2$. That turns the parabola into a lopsided W-like curve (but smoother than a W), shifted to the right of the origin, and touching the $x$-axis at -1.562…, 0, 0, and 2.562….
To find interesting points such as the minima and inflection points, though, you would need calculus. To give you a taste of what you haven't learned… we take the first derivative ("$\frac{d}{dx}\ f_1(x)$"), which tells you the slope of the curve for any $x$.
$$\frac{d}{dx}\ f_1(x) = 4x^3 - 3x^2 - 4 \cdot 2 \cdot x = \frac{x}{16} (8x - (3 - \sqrt{137}))(8x - (3 + \sqrt{137}))$$
We find that the slope is 0 at $x = 0$, $x \approx -1.088$, and $x \approx 1.838$. We conclude that $f_1(x)$ has a local maximum at (0, 0), a local minimum at (-1.088, -2.0458), and a local minimum at (1.838, -8.3097).
To find the inflection points, we would take the derivative of the derivative. This second derivative is
$$\frac{d^2}{dx^2}\ f_1(x) = 4 \cdot 3 \cdot x^2 - 3 \cdot 2 \cdot x - 8 = \frac{1}{12} (12x - (3 - \sqrt{105}))(12x - (3 + \sqrt{105}))$$
The inflection points, where the second derivative is zero, are at $x \approx -0.604$ and $x \approx 1.104$. That is, $f_1(x)$ is $\cap$-shaped in the interval $\frac{3-\sqrt{105}}{12} \lt x \lt \frac{3+\sqrt{105}}{12}$, and $\cup$-shaped everywhere else.
All of the information above can be seen in the plot. The roots are the red points, the extrema are green, and the inflection points are purple.