Forgot the closing parenthesis:
(2cos(t) + cos(2t))i + (2sin(t) + sin(2t))j + 0k
(in other words) : (2cos(t) + cos(2t))i + (2sin(t) + sin(2t)j + 0k
Compute the slope dy/dx and concavity d^2x/(dx^2) at t = pi/3.
I understand that finding the slope of a vector is possible by using dy/dx such that (dy/dt)/(dx/dt) and finding the concavity is by finding the derivative of (dy/dt) then divide (dy/dt)/(dx/dt) but what I'm confused here is when exactly do I plug t in? This part is really confusing to me and I would like some guidance on to actually solve for both the slope and concavity.
Any help is greatly appreciated!
(2cos(t) + cos(2t))i + (2sin(t) + sin(2t))j + 0k
Here's how you can calculate the slope dy/dx at t = π/3:
Step 1: Find dy/dt
Differentiate the y-component of the vector function with respect to t:
dy/dt = d/dt(2sin(t) + sin(2t))
= 2cos(t) + 2cos(2t)
Step 2: Find dx/dt
Differentiate the x-component of the vector function with respect to t:
dx/dt = d/dt(2cos(t) + cos(2t))
= -2sin(t) - 2sin(2t)
Step 3: Substitute t = π/3 into dy/dt and dx/dt
dy/dt = 2cos(Ï€/3) + 2cos(2Ï€/3) = 1 + (-1) = 0
dx/dt = -2sin(π/3) - 2sin(2π/3) = -√3 - √3 = -2√3
Step 4: Compute the slope dy/dx
dy/dx = (dy/dt) / (dx/dt) = 0 / (-2√3) = 0
So, the slope dy/dx at t = π/3 is 0.
To find the concavity d^2x/dt^2, you need to differentiate dy/dt with respect to t and then divide it by dx/dt.
Step 5: Find d^2y/dt^2
Differentiate dy/dt with respect to t:
d^2y/dt^2 = d/dt(2cos(t) + 2cos(2t))
= -2sin(t) - 4sin(2t)
Step 6: Find the concavity d^2x/dx^2
Substitute t = π/3 into d^2y/dt^2 and divide by dx/dt:
d^2x/dx^2 = (d^2y/dt^2) / (dx/dt) = (-2sin(π/3) - 4sin(2π/3)) / (-2√3)
= (-√3 - 4(-√3)) / (-2√3)
= (3 + 4√3) / (2√3)
So the concavity d^2x/dx^2 at t = π/3 is (3 + 4√3) / (2√3).
Remember to simplify your answers whenever possible.