Sine/Cosine Word Problem...Need Help...Please!?

Question

Deer jumps over fence
With graceful leaping beauty.
Sine curve models height.

Six feet above ground,
Each leap requires 1 second.
The equation, please?

When t=o
The deer is over the fence
At its maximum height.

I dont know what the period would be for this problem. Can someone help me find the sine and cosine equation for this. I don't know if it wants the cosine or sine equation. It would be nice if you showed the steps also. Thanks!

When you answer the question the equation needs pie in in and the midline.

Use the equation: y=Acos/sin(ktheta+c)+h

Answer 1

Very nicely put question.

I think it should be y = 3 cos (2pi t) + 3

This curve has the value 6 when t = 0.
The minimum height of the deer is 0 when it is on the ground
It says that it takes him one second to complete a leep, which
I understand to be from ground to ground again, so the period is 1. Hence the multiplication by the factor 2pi inside the cosine.

I disagree with the above answers but I could be wrong. It seems sensible to me to have minimum at y = 0.

Answer 2

This is a very beautiful problem, lady!

You may use both sine or cosine to describe the deer jump path. But the problem enunciation says sine, so, let's use sine instead of cosine!

You also may choose any coordinate system you want, but I believe that putting the origin on the floor, at base of the fence, with the y axis pointing to the sky and the x axis parallel to the floor, pointing at the same direction that the deer is going, makes it easier to understand.

The first thing you have to notice is:

"Sine curve models height".
That means the deer's height varies accordingly to a sine function of the time. So, what you are looking for is a function of the type:

y = A sin(tet) + B

Where "tet" is an angle that varies with the time (tet = V.t + fi) and A and B are constants.

The second thing you have to notice is not at the enunciation: when the deer jumps, it is at the floor (y=0) and it's vertical speed is maximum. When it is at it's maximum height, it is above the fence, and it's vertical speed is zero. When the deer lands, it is at the floor again (y=0), and it's vertical speed is maximum again. The function above has the same qualities at tet=0, tet=pi/2 and tet=pi rad! So we can use them!

Let's make:

At tet=0, y = 0;
At tet=pi/2, y=6.
("Six feet above ground, ... ").

So,

0 = A sin(0) + B
=> B = 0 ft

6 = A sin(pi/2) + B
=> A = 6 ft

Our function becomes:
y = 6 sin(tet)

And let's not forget that:

"When t = 0
The deer is over the fence
At its maximum height"

So at t = 0, tet=pi/2

But:

"Six feet above ground,
Each leap requires 1 second."

By reading this, I understood that each complete leap (jumping, passing through the fence and landing) requires 1 second. So the deer takes 1/2s to reach the floor after it passes the fence:

At t=1/2, tet=pi

Solving tet = V.t + fi :

pi/2 = V.0 + fi => fi = pi/2 rad
pi = V.1/2 + fi => V = pi rad/s

So tet = pi.t + pi/2

and y = 6 sin(pi.t + pi/2)

And let's not forget that sine values range from -1 to 1. We are not interested in negative numbers, for the deer is not going to walk under ground. This problem may be solved putting the sine function inside a modulus function:

y = mod[ 6 sin(pi.t + pi/2)]

This also make sure that the deer keep jumping after it jumps the fence, as long as you give time to it! ... :-)

You would be tempted to say that the period of this function is 2s since it takes 2s for "tet" run 2pi rad, and 2pi rad is the period of the sine function (the period of a function, say y=f(x), is the range that x runs to make y values repeat. 2pi rad is the range of sine and cosine functions, because after 2pi rad the sine or cosine are the same).

BUT we put our function inside a modulus function, making negative values of the sine equal the positive ones! So "tet" has to run only pi rad to our sine function repeat its values. Knowing that it takes 1s to "tet" run pi rad, the period of our function is 1s!!

Answer 3

6cos(pi*t) so at t=1/2 it is back on the ground, the first half of the jump from ground to max height is 0.5 seconds and then back to the ground in 0.5secs. This is in radians. In degrees the answer would be
6cos(180t)

Answer 4

y=6cost