Question
Find f.
f ''(θ) = sin θ + cos θ, f(0) = 5, f '(0) = 3
My steps:
f'(θ)=cosθ-sinθ+C
When f'(0)=3, C=-2, so f'(θ)=cosθ-sinθ-2.
f(θ)=-sinθ-cosθ-2x+D
When f(0)=5, D=6, so f is -sinθ-cosθ-2x+6.
How is that wrong?
f ''(θ) = sin θ + cos θ, f(0) = 5, f '(0) = 3
My steps:
f'(θ)=cosθ-sinθ+C
When f'(0)=3, C=-2, so f'(θ)=cosθ-sinθ-2.
f(θ)=-sinθ-cosθ-2x+D
When f(0)=5, D=6, so f is -sinθ-cosθ-2x+6.
How is that wrong?
Answers
Reiny
f'Ø) = -cosx + sinx + C
you differentiated instead of integrated.
Had you checked your answer by taking the derivative you would have seen your error
since f'(0) = 3
3 = -cos0 + sin0 + c
3 = -1 + 0 + c ----> c = 4
so f'(x) = -cosx + sinx + 4
f(x) = -sinx - cosx + 4x + d
and f(0) = 5
5 = -sin0 - cos0 + 0 + d
5 = 0 - 1 + d
d = 6
f(x) = -sinx - cosx + 4x + 6
Just realized I used x instead of Ø, no big deal to change
I suggest you take a better look at the derivatives of both sine and cosine
you differentiated instead of integrated.
Had you checked your answer by taking the derivative you would have seen your error
since f'(0) = 3
3 = -cos0 + sin0 + c
3 = -1 + 0 + c ----> c = 4
so f'(x) = -cosx + sinx + 4
f(x) = -sinx - cosx + 4x + d
and f(0) = 5
5 = -sin0 - cos0 + 0 + d
5 = 0 - 1 + d
d = 6
f(x) = -sinx - cosx + 4x + 6
Just realized I used x instead of Ø, no big deal to change
I suggest you take a better look at the derivatives of both sine and cosine