
#1
May2811, 10:35 PM

P: 67

Hi
A weight hangs on the end of an elastic string. The weight is pulled down and released from rest at time t = 0. The length x centimeters of the string at time t seconds is given by the formula: x = 7  2sin(πt) a) Sketch the graph of x against t for 0 <= x <=4 b) Find the two times at which the length of the string is least and the values of x at those times c) After how many seconds does the weight return to its starting point for the first time? My Answers b) So, I started out by thinking about what would be the least length for the function sin(x). I got to the conclusion it was 1, which is equal to an angle of π/2 radians or 1.5π radians. Applying the angle in the formula gave me 9 while the correct answer is 5. The answer 5 is obtained when using sin(1) instead of sin(1). Why? Furthermore, the question asks for the times at which the value of x is the least before the value of x itself. Is it possible to solve it in that order, because, what I did next was: 5 = 7  2sin(πt) and solved for t, finding 0.5. But I don't know how to find the second answer and I don't even know where to start for letter C... Any help appreciated. 



#2
May2811, 10:43 PM

P: 1,035

When you look at the function, you know right off the bat what the min and max values are going to be. In the form asin(bx+c)+d, a is the amplitude, so it's going up and down 2 units. You will also notice that D in your function is +7, so you know the whole thing is shifted up 7 from the X axis. If the wave makes an arc that has a height 2 from it's axis, then makes an arc that has a height 2, the max and min can be given by 7+2 and 72.
From there, you can just set your equation equal to those two values and solve for all solutions of the trig equation, then find the values (coterminals or "other angles" whatever you call them) that are within the range the question asks. I made an image because I am bored: 



#3
May2811, 11:09 PM

P: 67

Ok cool, thanks for that.
So, I tried letter C now: The rest position is 0. We know it crosses 0 the first time at 0 seconds. We need to know what will be the time when the function will cross 0 again. To do so, I did 180  my angle, hence 180 degrees and converted it to radians, getting π as an answer. Is that correct? Thanks 



#4
May2811, 11:13 PM

P: 1,035

Trigonometric Equations
What you need to do is start from those two equations in the above image, and then whittle it down until you get the general solution for the angle t. In the general solution you will have a variable n (sometimes called k) that determines how many rotations get you to the same answer basically. You need to input a number for that, so it satisfies the questions parameters of "the second time it reaches 0 seconds".




#5
May2911, 08:20 AM

P: 67

Thanks for the diagram and help. I understand what you are saying but there's still one thing I don't get. If you draw the original sin(x) graph and apply the transformations from the lowest point in the graph 1 you get to the highest value. Is that because this graph has a reflection in the xaxis?




#6
May2911, 02:27 PM

P: 1,035

The k value, transforms it, shifting up and down the Y axis. It starts from 0. That gives you the new "axis" which the standard graph of sine can be drawn on. From that axis, up/down A(amplitude) will give you the min and max. The negative amplitude reverses which direction (up/down) the wave moves first.
Take a look at the dotted line wave in the image. That is the normal graph of sine. Notice it reaches it's max before it's min. 


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