Pre-calculus HELP>>>>>>>>>>>>>?

Question

Solve the problem: A mass hangs from a spring which oscillates up and down. the Postition P (in feet) of the mass at time t(inseconds) is given by=P=4cos(4t)
For what values of t will the position be 2 square ROOT of 2? Find the exact value..give the ans in terms of pie PLEASE SHOW YOUR WORK..THANKYOU

Answer 1

This is asking you to solve the equation
4*cos(4t) = 2 √2
cos(4t) = √2 / 2

You should know from trigonometry that cosine is √2 / 2 at pi/4+2k*pi and -pi/4+2k*pi, for an integer k. Thus you can simplify the problem to
4t = pi/4+2k*pi OR 4t = -pi/4+2k*pi
t = pi/16+1/2*k*pi OR t = -pi/16+1/2*k*pi

Note that this equation is in general a poor approximation of an actual spring. Its motion is much better modeled by differential equations, which you can look forward to in college!

Cheers,
Nicholas

Answer 2

First, you have to figure out what form of 4cosine will give you the answer of (2 square root of 2). You do that by dividing P by 4 obviously. With the left part of the equation (2 square root of 2)/4, you can divide because there's a multiple of 2 in both the denominator and the numerator. So, you divide by 2...giving you "(square root of 2)/2". Knowing this information, you should use the Pythagorean theorem of a^2 + b^2= c^2. So thus, in the unit circle, you know c^2=1. Therefore, if you square (square root of 2)/2, you get 2/4 --> 1/2. Therefore, its another square root of 2/2 as the other side. Short answer, it has to be 45 degrees=4t, but that's in degrees. If you're using radians (pie), then 4t has to be equal to pie/4. If it's FOUR t, however, then you divide by four and get pie/16. Hope that's helpful.

It might have seemed weird/complicated but if you need to show your work, then you should try to incorporate those steps as best possible.

Good luck in Pre-Calc.

Answer 3

P = 2√2 = 4cos(4t)
cos(4t) = (1/2)√2
4t = π/4 + 2πn
t = π/16 + (π/2)n
or
4t = 7π/4 + 2πn
t = 7π/16 + (1/2)πn

so the times go like
π/16, 7π/16, 9π/16, 15π/16, ....

Answer 4

okay, we need p to equal two times the square root of two.

2(square root of two)=4cos(4t)

square root of two=2cos(4t)

square root of two/2=cos(4t)

cos^-1(square root of two/2)=4t

t=(cos^-1(square root of two/2))/4

t=cos^-1(.707106...)/4

t=.785399.../4

t=.19634...

to express in terms of pi, simply factor out pi

.19634/3.14159265423...

t=.062... times pi


I think this is right. Good luck. This seems too easy to be precalculus.

Answer 5

first it is pi not pie. And try to write correctly your question.
Too many mistakes.

then you have to solve the equation

P(t)=4 cos(4t) = 2sqrt(2)

cos(4t) = sqrt(2)/2
Now look on your textbook for which values cos is sqrt(2)/2

Answer 6

2root2=4cos(4t)
root2/2=cos(4t)
4t= pi/4 (from the unit circle- cos(pi/4)=root2/2)
16t=pi
t=pi/16

Answer 7

2*SQRT(2)=4cos(4t)
SQRT(2)/2=cos(4t)
arccos(SQRT(2)/2)=4t
PI/4=4t OR 7PI/4=4t OR 9PI/4=4t OR 15PI/4 ETC
PI/16=t OR 7PI/16 OR 9PI/16=t OR 15PI/16=t
Etc, etc...