A body is moving in simple harmonic motion with position function s(t)=2+2cost where s is in meters and t is in seconds. Find the jerk at time t. (1 point) Responses j(t)=0 j ( t ) = 0 j(t)=−2cost j ( t ) = − 2 cos t j(t)=2cost j ( t ) = 2 cos t j(t)=−2sint j ( t ) = − 2 sin t j(t)=2sint
Jerk is defined as the time derivative of acceleration. In this case, we need to find the second derivative of the position function in order to find the acceleration, and then take the derivative of acceleration to find the jerk.
Given position function:
s(t) = 2 + 2cos(t)
First, let's find the first derivative (velocity):
v(t) = d/dt [s(t)]
v(t) = -2sin(t)
Next, let's find the second derivative (acceleration):
a(t) = d/dt [v(t)]
a(t) = d/dt [-2sin(t)]
a(t) = -2cos(t)
Finally, let's find the derivative of acceleration (jerk):
j(t) = d/dt [a(t)]
j(t) = d/dt [-2cos(t)]
j(t) = 2sin(t)
Therefore, the jerk at time t is j(t) = 2sin(t).
To find the jerk at a given time, we need to take the third derivative of the position function.
Given that the position function is s(t) = 2 + 2cos(t), we can start by finding the derivatives:
First derivative: s'(t) = -2sin(t)
Second derivative: s''(t) = -2cos(t)
Third derivative (Jerk): j(t) = 2sin(t)
Therefore, the jerk at time t is j(t) = 2sin(t).
To find the jerk at time t, we need to differentiate the position function twice.
Given:
s(t) = 2 + 2cos(t)
First, we differentiate s(t) with respect to time (t):
v(t) = d/dt [s(t)]
= d/dt [2 + 2cos(t)]
= -2sin(t)
Next, we differentiate v(t) with respect to time (t) to find the acceleration:
a(t) = d/dt [v(t)]
= d/dt [-2sin(t)]
= -2cos(t)
Finally, we differentiate a(t) with respect to time (t) to find the jerk:
j(t) = d/dt [a(t)]
= d/dt [-2cos(t)]
= 2sin(t)
Therefore, the jerk at time t is given by:
j(t) = 2sin(t)
Find the equation for the lines that are tangent and normal to y=sinx−3 at x = π. (1 point) Responses tangent: y=−2 ; normal: x = π tangent: y = − 2 ; normal: x = π tangent: y=−4 ; normal: x = π tangent: y = − 4 ; normal: x = π tangent: y=−x+π−3 ; normal: y=−x−π−3 tangent: y = − x + π − 3 ; normal: y = − x − π − 3 tangent: y=−x+π−3 ; normal: y=x−π−3 tangent: y = − x + π − 3 ; normal: y = x − π − 3 tangent: y=−x−π−3 ; normal: y=x−π−3
To find the equation for the tangent line, we can find the derivative of the given function, y = sin(x) - 3:
dy/dx = cos(x)
At x = π, the derivative is cos(π) = -1.
Therefore, the slope of the tangent line is -1.
Now, using the point-slope form of a line, with the point (π, sin(π) - 3), we can write the equation for the tangent line:
y - (sin(π) - 3) = -1(x - π)
Simplifying, we get:
y = -x + π - 3
So, the equation for the tangent line is y = -x + π - 3.
To find the equation for the normal line, we need to find the negative reciprocal of the slope of the tangent line.
The negative reciprocal of -1 is 1.
Again, using the point-slope form with the point (π, sin(π) - 3), we can write the equation for the normal line:
y - (sin(π) - 3) = 1(x - π)
Simplifying, we get:
y = x - π - 3
So, the equation for the normal line is y = x - π - 3.
Therefore, the equations for the tangent and normal lines are:
Tangent: y = -x + π - 3
Normal: y = x - π - 3
Find the derivative of 5sinx . (1 point) Responses 5cosx 5 cos x −5cotxcscx − 5 cot x csc x 5cotxcscx 5 cot x csc x 5cosx 5 cos x 5cotxcscx−sec2x 5 cot x csc x − sec 2 x
The derivative of 5sinx can be found by applying the chain rule.
Using the chain rule, the derivative of sin(x) is cos(x), and multiplying it by the derivative of the function inside the sine (which is 5), we get:
d/dx (5sinx) = 5 * cosx = 5cosx
Therefore, the derivative of 5sinx is 5cosx.