# Thread: Anyone smarter than me! You're needed!

1. ## Anyone smarter than me! You're needed!

Whoever could do that would be (imo) an absolute legend!!!!

2. Mmmk. This is way too much for me personally to give you all the answers. Some other members might, but I'll get you started.

For #1 I am confused on some notation. Is it $D(t)=5+3\cos(\frac{\pi}{6})(t+2)$ or $D(t)=5+3\cos(\frac{\pi}{6}(t+2))$?

Either way, for A)i. plug in D(0). For ii. and iii. take the derivative of the function to get the max/min values.

3. Originally Posted by Iskarius
Whoever could do that would be (imo) an absolute legend!!!!

...
Hi, Iskarius,

I presume that you mean: $D(t)=5+3\cos(\frac{\pi}{6}(t+2))$

to Aii), iii) The values of the Cosine-function are between -1 and 1. Therefore the depth can only between 2 m and 8 m.

I've attached a diagram to do F).

Bye

EB

4. Hello, Iskarius!

A jet of water in a backyard is of parabolic shape.
It starts 3 metres above the ground and just passes over the top of a tree
that is 5 metres high and is a distance of 2 metres horizontally from the starting point.
The jet of water strikes the surface of a flower bed at ground level
at a distance of 8 metres horizontally from the starting point.

A) Which of the following equations (in which a, b, c are constants) models the jet of water?
$(1)\;y = (x-a)(x-b)(x-c)\qquad(2)$ $y = x(x-a)(x-b) + c$
$(3)\;y = x^3 + ax^2 + bx + c\qquad(4)\;a(x-b) + c\qquad(5)$ $ax^2 + bx + c$
The path is a parabola, hence its equation is a quadratic (second-degree).

Since $(1),\;(2),\;(3)$ are cubic functions and $(4)$ is a straight line,
. . the answer is the only quadratic: . $(5)\;ax^2 + bx + c$

B) Use answer from part A to determine the values of a, b, and c.
Hence, state the equation of the path traced out by the jet of water.
Place the base of the tree at the origin and the graph looks like this:
Code:
              |   *
(0,5)o       *
*  |          *
*    |            *
(-2,3)o     |             *
|     |
--+-----+--------------o--
-2     0            (6,0)
The general equation is: . $y \:=\:ax^2 + bx + c$

Since $(0,5)$ is on the graph, we have:
. . $a\cdot0^2 + b\cdot0 + c \:=\:5\quad\Rightarrow\quad c = 5$

The function (so far) is: . $y\:=\:ax^2 + bx + 5$

Since $(-2,3)$ is on the graph, we have:
. . $a(-2)^2 + b(-2) + 5\:=\:3\quad\Rightarrow\quad 4a - 2b\:=\:-2$

Since $(6,0)$ is on the graph, we have:
. . $a\cdot6^2 + b\cdot6 + 5\:=\:0\quad\Rightarrow\quad 36a + 6b \:= \:-5$

Solve this system of equation and we get: . $a = -\frac{11}{48},\;\;b = \frac{13}{24}$

Therefore, the equation is: . $\boxed{y \:= \:-\frac{11}{48}x^2 + \frac{13}{24}x + 5}$

C) Find out how far the jet of water rises above the ground.
The graph is a down-opening parabola; its highest point is at its vertex.

The vertex of the parabola $y \:=\;ax^2 + bx + c$ is at: . $x = \frac{-b}{2x}$

Our parabola has: $a = -\frac{11}{48},\;b = \frac{13}{24}$

Its vertex is at: . $x \:=\:\frac{-\frac{13}{24}}{2\left(-\frac{11}{48}\right)}} \:= \:\frac{13}{11}$

Therefore, the maximum height is:
. . $y \:=\:-\frac{3}{8}\left(\frac{13}{11}\right)^2 + \frac{1}{4}\left(\frac{13}{11}\right) + 5 \:=\:\frac{1489}{264} \:\approx \:\boxed{5.64}$ metres.

D) Find the angle of the jet to the horizontal at the starting point.
We need Calculus for this part . . .

The derivative is: . $y' \:=\:-\frac{11}{24}x + \frac{13}{24}$

At $x = -2$, the slope of the tangent line is: . $m \:=\:-\frac{11}{24}(-2) + \frac{13}{24} \:= \:\frac{35}{24}$

Therefore: . $\tan\theta \:= \:\frac{35}{24}\quad\Rightarrow\quad\theta \:= \:\arctan\left(\frac{35}{24}\right) \:\approx\:\boxed{55.56^o}$