If f is a vector-valued function defined by <sin2t, cos2t> then what is f '' (pi/4)?
I think it is <-4, -√2/2>.
Sorry...it is cost, not cos2t.
x= sin 2 t
y = cos t
x' = 2 cos 2t
y' = -sin t
x" = - 4 sin 2t
y" = -cos t
at pi/4
sin pi/2 = 1
cos pi/4 = sqrt 2/2
so yes , (-4 , -sqrt 2 / 2 )
To find the second derivative of a vector-valued function, you need to differentiate each component of the function twice.
Let's start by finding the first derivative of the vector-valued function f(t).
f(t) = <sin(2t), cos(2t)>
To find f'(t), we differentiate each component of f(t) with respect to t:
f'(t) = <d/dt(sin(2t)), d/dt(cos(2t))>
Using the chain rule, we have:
f'(t) = <2cos(2t), -2sin(2t)>
Now let's find the second derivative of the vector-valued function f(t). We differentiate each component of f'(t) with respect to t:
f''(t) = <d/dt(2cos(2t)), d/dt(-2sin(2t))>
Applying the chain rule again, we get:
f''(t) = <-4sin(2t), -4cos(2t)>
Now we can substitute t = π/4 into f''(t) to find f''(π/4):
f''(π/4) = <-4sin(2(π/4)), -4cos(2(π/4))>
Simplifying further:
f''(π/4) = <-4sin(π/2), -4cos(π/2)>
Since sin(π/2) = 1 and cos(π/2) = 0:
f''(π/4) = <-4(1), -4(0)>
f''(π/4) = <-4, 0>
Therefore, the second derivative of the vector-valued function f at t = π/4 is <-4, 0>.
So your initial guess of <-4, -√2/2> is incorrect. The correct answer is <-4, 0>.