The endpoints of a movable rod of length 1 m have coordinates (x, 0) and (0, y) (see figure). The position of the end on the x-axis is given in the equation below, where t is the time in seconds.
x(t) = 1/8 sin(πt/4)
(c) Find the speed of the y-axis endpoint when the x-axis endpoint is (1/16, 0). (Round your answer to four decimal places.)
Well, let's break it down step by step. To find the speed of the y-axis endpoint, we need to find the derivative of y with respect to time.
Since we know that the x-coordinate of the endpoint is given by x(t) = (1/8)sin(πt/4), we can use this to find the equation for y-coordinate.
Since the length of the rod is 1m, we can use the Pythagorean theorem to find y in terms of x:
y^2 = (1 - x)^2
y = √((1 - x)^2)
Now, we can differentiate y with respect to t to find the speed:
dy/dt = d/dt[√((1 - x)^2)]
To simplify this, let's use the chain rule:
dy/dt = (1/2)[(1 - x)^2]^(-1/2)(-1)(dx/dt)
Since we know that dx/dt = (π/32)cos(πt/4), we can substitute this back into the equation:
dy/dt = (1/2)[(1 - x)^2]^(-1/2)(-1)(π/32)cos(πt/4)
Now, let's substitute the given x-coordinate, x = 1/16, into the equation:
dy/dt = (1/2)[(1 - 1/16)^2]^(-1/2)(-1)(π/32)cos(πt/4)
Simplifying further:
dy/dt = (1/2)[(15/16)^2]^(-1/2)(-1)(π/32)cos(πt/4)
Now, we can calculate the speed by evaluating dy/dt at the given time. Since the time is not provided, I'm afraid we can't give you an exact answer. But hey, at least we went through all the math!
To find the speed of the y-axis endpoint, we need to find the derivative of the y-coordinate with respect to time (t).
Given that the x-coordinate is given by the equation x(t) = (1/8) sin(πt/4), we can find the y-coordinate using the Pythagorean theorem.
The length of the rod (l) is given by: l = sqrt(x^2 + y^2)
Substituting the given x-coordinate (1/16) into the equation, we have:
l = sqrt((1/16)^2 + y^2)
Simplifying, we get:
l = sqrt(1/256 + y^2)
Since we know that the length of the rod is 1 m, we can set l equal to 1 and solve for y:
1 = sqrt(1/256 + y^2)
Squaring both sides, we have:
1 = 1/256 + y^2
Rearranging the equation, we get:
y^2 = 1 - 1/256
y^2 = 255/256
Taking the square root of both sides, we have:
y = sqrt(255/256)
Now, to find the speed of the y-coordinate, we differentiate y with respect to t:
y'(t) = d/dt(sqrt(255/256))
Using the chain rule, we get:
y'(t) = (1/2)(-1/2)(-1)(255/256)^(-1/2)(d/dt(255/256))
y'(t) = (1/2)(-1/2)(-1)(255/256)^(-1/2)(0)
y'(t) = 0
Therefore, the speed of the y-axis endpoint when the x-axis endpoint is (1/16, 0) is 0.
To find the speed of the y-axis endpoint when the x-axis endpoint is (1/16, 0), we need to differentiate the position equation with respect to time to get the velocity equation.
Given:
x(t) = 1/8 sin(πt/4)
To find y(t), we need to substitute the given x-coordinate (1/16) into the equation:
1/16 = 1/8 sin(πt/4)
Now, we can solve this equation for t. Let's rearrange the equation:
sin(πt/4) = (1/16) * 8
sin(πt/4) = 1/2
To find t, we will use the inverse sine function (sin^(-1)):
πt/4 = sin^(-1)(1/2)
Now, let's solve for t:
πt = 4 * sin^(-1)(1/2)
t = (4 * sin^(-1)(1/2)) / π
Now that we have the value of t at the given x-coordinate, we can find y(t):
y(t) = 0 - y-coordinate of the x-axis endpoint = 0
So, the y-coordinate is 0.
Now, let's differentiate x(t) with respect to t to get the velocity equation:
dx/dt = (1/8) * (π/4) * cos(πt/4)
The velocity equation for x is:
v_x(t) = (π/32) * cos(πt/4)
To find the speed of the y-axis endpoint, we need to find the magnitude of the velocity vector, which is the square root of the sum of the squares of the x and y components of the velocity.
Since the y-component of the velocity is zero (dy/dt = 0), the speed of the y-axis endpoint is simply given by the magnitude of the x-component of the velocity:
v_y(t) = v_x(t) = (π/32) * cos(πt/4)
Now, let's substitute the value of t we found earlier:
t = (4 * sin^(-1)(1/2)) / π
v_y(t) = (π/32) * cos((π/4) * (4 * sin^(-1)(1/2)) / π)
Simplifying the expression, we have:
v_y(t) = (π/32) * cos(sin^(-1)(1/2))
Using the identity cos(sin^(-1)(x)) = √(1 - x^2), we can further simplify:
v_y(t) = (π/32) * √(1 - (1/2)^2)
Simplifying further:
v_y(t) = (π/32) * √(3/4)
Finally, calculating the value:
v_y(t) = (π/32) * √3 ≈ 0.0547 (rounded to four decimal places)
Therefore, the speed of the y-axis endpoint when the x-axis endpoint is (1/16, 0) is approximately 0.0547.
sketch it.
length= L = 1m
1= x^2+y^2
0= 2x dx/dt + 2y dy/dt
dy/dt= -(x/y)dx/dt
but x=1/2 sin PIt/4
dx/dt= 1/2* cos(PIt/4)*PI/4
Now when x=1/16, then y= sqrt (L^2-x^2)
at x=1/16, then y= sqrt(1-1/256)
y= sqrt (255/256)
dx/dt= PI/8 * cos (PIt/4)
and x/y= (1/16)/(255/256)=16/255
dy/dt=-(x/y)dx/dt= you do it.
check my work, I am tired tonite.