h(t) 0.6cos(2t) + 08 sin(t)
h'(t) = v(t) = -1.2sin(2t) + 0.8cos(t)
find the vertical displacement when the velocity is 0.8m/s
how do i solve v(t) = 0.8m/s
I tried the normal way by equating and factoring, however two answers are obtained and the second method is the identity: asin(x) + bcos(x) = Rsin(x+theeta). In the identity to find theeta should i consider the -1.2 or just its absolute value? because then i get t=0 which is the only answer.
Perhaps there is an easier way, but this is what I suggest for now:
h'(t) = v(t) = -1.2sin(2t) + 0.8cos(t) =0.8
-1.2sin(2t) + 0.8cos(t) = 0.8
Divide by 0.8 throughout to get
-1.5sin(2t) + cos(t) = 1
expand sin(2t) by the double angle formula and factor out cos(t)
-3sin(t)cos(t) + cos(t) = 1
cos(t)(1-3sin(t)) = 1
Substitute c=cos(t) and √(1-c²)=sin(t):
c(1-3√(1-c²)) = 1
square (and verify roots later)
(c-1)²=9c²(1-c²)
We know that c=1 is a factor, so do a long division and solve the resulting cubic to get
c1=1, c2=(sqrt(19)/3^(9/2)+1/27)^(1/3)+2/(27*(sqrt(19)/3^(9/2)+1/27)^(1/3))-1/3=0.25643078474621
or you can solve for c2 numerically.
The other two roots are complex, so they are rejected automatically.
Now solve for t:
t1=acos(c1)=acos(1) = 0
t2=±acos(c2)=±acos(0.25643078474621)
=±1.311468632085831
substitute into v(t):
v(0)=0.8
v(1.311468632085831)=-0.3897, rejected.
v(-1.311468632085831) = 0.8 OK
To get a positive value for t2, add 2π to get
t2=4.971716675093755
To solve the equation v(t) = 0.8 m/s, which is the equation for velocity, we have the equation:
-1.2sin(2t) + 0.8cos(t) = 0.8
To solve this equation, you can rearrange terms and use trigonometric identities. Let's start by isolating the sine term:
-1.2sin(2t) = 0.8 - 0.8cos(t)
Now, divide both sides by -1.2:
sin(2t) = (0.8 - 0.8cos(t)) / -1.2
Next, let's use the identity you mentioned: asin(x) + bcos(x) = Rsin(x + θ). In this case, a = 0, b = -0.8, and R = 1.2. To find θ, we can consider the coefficient of sin(t+θ), which is -1.2 in our case. Since the coefficient is negative, we have:
sin(2t) = 1.2sin(t + θ)
Now, let's compare the arguments of sine on both sides. We have:
2t = t + θ
Simplifying the equation, we get:
t = θ
So, the value of t will be equal to the value of θ. In this case, the solution for t is also the value for θ.
Now, substituting θ back into the equation, we have:
sin(2t) = 1.2sin(t)
To solve this equation, we can apply another trigonometric identity, sin(2t) = 2sin(t)cos(t):
2sin(t)cos(t) = 1.2sin(t)
Divide both sides by sin(t):
2cos(t) = 1.2
Finally, divide both sides by 2:
cos(t) = 0.6
To find the value of t, you can use the inverse cosine function (also known as arccosine), which gives the angle whose cosine is equal to 0.6. So, taking the inverse cosine of both sides, we get:
t = arccos(0.6)
Using a calculator or a table of trigonometric values, you can find the value of arccos(0.6) to be approximately 0.9273 radians or approximately 53.1301 degrees.
Therefore, the solution for t is t = 0.9273 radians or t ≈ 53.1301 degrees.
Keep in mind that there might be other possible solutions, as trigonometric functions have periodic behavior. However, in this specific case, the solution t = 0.9273 radians is the one that satisfies the equation v(t) = 0.8 m/s.
As for the vertical displacement, you can substitute this value of t back into the original function h(t) = 0.6cos(2t) + 0.8sin(t) to find the corresponding displacement h(t) at that time.