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

http://i2.photobucket.com/albums/y17/Nath015/Math2.jpg

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- Jun 28th 2006, 10:21 PMIskariusAnyone smarter than me! You're needed!
Whoever could do that would be (imo) an absolute legend!!!!

http://i2.photobucket.com/albums/y17/Nath015/Math2.jpg - Jun 29th 2006, 08:19 AMJameson
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 $\displaystyle D(t)=5+3\cos(\frac{\pi}{6})(t+2)$ or $\displaystyle D(t)=5+3\cos(\frac{\pi}{6}(t+2))$?

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. - Jun 29th 2006, 08:29 AMearbothQuote:

Originally Posted by**Iskarius**

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

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 - Jun 29th 2006, 12:49 PMSoroban
Hello, Iskarius!

Quote:

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?

$\displaystyle (1)\;y = (x-a)(x-b)(x-c)\qquad(2)$ $\displaystyle y = x(x-a)(x-b) + c$

$\displaystyle (3)\;y = x^3 + ax^2 + bx + c\qquad(4)\;a(x-b) + c\qquad(5)$$\displaystyle ax^2 + bx + c$

Since $\displaystyle (1),\;(2),\;(3)$ are cubic functions and $\displaystyle (4)$ is a straight line,

. . the answer is the only quadratic: .$\displaystyle (5)\;ax^2 + bx + c$

Quote:

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.

Code:`| *`

(0,5)o *

* | *

* | *

(-2,3)o | *

| |

--+-----+--------------o--

-2 0 (6,0)

Since $\displaystyle (0,5)$ is on the graph, we have:

. . $\displaystyle a\cdot0^2 + b\cdot0 + c \:=\:5\quad\Rightarrow\quad c = 5$

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

Since $\displaystyle (-2,3)$ is on the graph, we have:

. . $\displaystyle a(-2)^2 + b(-2) + 5\:=\:3\quad\Rightarrow\quad 4a - 2b\:=\:-2$

Since $\displaystyle (6,0)$ is on the graph, we have:

. . $\displaystyle a\cdot6^2 + b\cdot6 + 5\:=\:0\quad\Rightarrow\quad 36a + 6b \:= \:-5$

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

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

Quote:

C) Find out how far the jet of water rises above the ground.

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

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

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

Therefore, the maximum height is:

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

Quote:

D) Find the angle of the jet to the horizontal at the starting point.

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

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

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