# Math Help - Simultaneous equations

1. ## Simultaneous equations

Hi,

I have a maths exam tomorrow, so I would be grateful for any help!

So here it is:

Let f(t) = a cos b(t-c) + d (I understand that this refers to the transformation)

when t = 3, there is a maximum value of 29,

when t = 9, there is a minimum value of 15.

Find the value of a.

I got as far as:
29 = a cos 3t
15 = a cos 9t

Is that right? And how should I find a?

Thanks!

2. Hi

The maximum possible value for a cosines is 1 and the minimum is -1
Therefore
a + d = 29
-a + d = 15
which leads to a = 7 and d = 22

3. Originally Posted by CSG18
Hi,

I have a maths exam tomorrow, so I would be grateful for any help!

So here it is:

Let f(t) = a cos b(t-c) + d (I understand that this refers to the transformation)

when t = 3, there is a maximum value of 29,

when t = 9, there is a minimum value of 15.

Find the value of a.

I got as far as:
29 = a cos 3t
15 = a cos 9t

Is that right? And how should I find a?

Thanks!
No, I don't think that's right. Remember that $\cos$ assumes values in the range $[-1;+1]$. If you assume that $a>0$, you get that the largest value of $f(t)$ is going to be $a+d$ and the smallest value is going to be $-a+d$. Hence a and d have to satisfy two equations: $29=a+d$ and $15=-a+d$, from which you get the correct values for a and d.

Next you must understand that this problem has infinitely many solutions. This is because, the distance $9-3=6$ between the t coordinates of a minimum and a maximum must be half the period (plus, optionally, an integral multiple of the period) of $f(t)$. This allows you to determine the (infinitely many possible) value(s) of $b$, because the period of $f(t)$ satisfies the relation $T=\frac{2\pi}{b}$.
Given the right value for $b$ you can now choose the correct value for $c$ by noting that you just have to shift the the graph $y=a\cos(bt)+d$ by 3 units to the right in the x-direction.

4. Thank you very much! Although, I didn't understand by what you meant when you said: "you can now choose the correct value for c"

But thanks again!

5. Originally Posted by CSG18
Thank you very much! Although, I didn't understand by what you meant when you said: "you can now choose the correct value for c"

But thanks again!
Suppose you have the graph $y=g(t)$, and you want to shift this graph by 3 units to the left. What do you do? - Well, you just replace every occurrence of t in $g(t)$ by $t-3$. That's it.

6. Oh, so in my original question, if you wanted to find the value of C, you would substitute say, the maximum data into the f(t) equation?

So as f(t) = 7 cos b (3-c) + d

would c = 3, because 3-3 = 0?

7. Originally Posted by CSG18
Oh, so in my original question, if you wanted to find the value of C, you would substitute say, the maximum data into the f(t) equation?

So as f(t) = 7 cos b (3-c) + d

would c = 3, because 3-3 = 0?
Sure: for the argument $b(3-3)=0$, you have that $\cos$ assumes its maximum value, $+1$, as required for $t=3$.