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?
Well, it seems like you're on the right track with finding the second derivative and integrating it to get the general form of the function. However, I can't help but feel like there's something a bit off about your approach. Maybe it just needs a little clowning around to set things straight!
Let's take a step back and try a different approach, shall we?
Since we are given f''(θ) = sin(θ) + cos(θ), we can integrate it to find f'(θ):
f'(θ) = -cos(θ) + sin(θ) + C1
Now, we can use the given condition f'(0) = 3 to determine C1. Plugging in θ = 0:
3 = -cos(0) + sin(0) + C1
3 = -1 + 0 + C1
C1 = 4
Great! Now that we have f'(θ) = -cos(θ) + sin(θ) + 4, we can integrate it again to find f(θ):
f(θ) = -sin(θ) - cos(θ) + 4θ + C2
To determine C2, we can use the given condition f(0) = 5. Plugging in θ = 0:
5 = -sin(0) - cos(0) + 0 + C2
5 = -0 - 1 + 0 + C2
C2 = 6
Voila! Our final function is f(θ) = -sin(θ) - cos(θ) + 4θ + 6. So it seems like your original answer was pretty close, but just a few missteps along the way. Don't worry, even the best clowns trip up occasionally!
Your steps are not entirely correct. Let's go through the correct steps to find f(θ):
Step 1: Find the antiderivative of f''(θ):
Integrating f''(θ) = sinθ + cosθ, we get:
f'(θ) = -cosθ - sinθ + C1,
Step 2: Use the initial condition f'(0) = 3 to solve for C1:
f'(0) = -cos(0) - sin(0) + C1 = -1 + C1 = 3,
C1 = 4.
Step 3: Find the antiderivative of f'(θ):
Integrating f'(θ) = -cosθ - sinθ + 4, we get:
f(θ) = -sinθ - cosθ + 4θ + C2.
Step 4: Use the initial condition f(0) = 5 to solve for C2:
f(0) = -sin(0) - cos(0) + 4(0) + C2 = -1 + C2 = 5,
C2 = 6.
So, the final solution for f(θ) is:
f(θ) = -sinθ - cosθ + 4θ + 6.
Your steps seem mostly correct, but there is a small mistake in the final expression for f(θ). Let's go through the steps together:
Given: f''(θ) = sin(θ) + cos(θ), f(0) = 5, f'(0) = 3.
Step 1: Solve for f'(θ)
To find f'(θ), we integrate the given f''(θ) with respect to θ:
f'(θ) = ∫ (sin(θ) + cos(θ)) dθ
Using the integral rules, we get:
f'(θ) = -cos(θ) + sin(θ) + C
Step 2: Finding the constant of integration (C)
Substituting θ = 0 into f'(θ) and equating it to the given f'(0) = 3, we have:
-1 + 0 + C = 3
C = 4
So, f'(θ) = -cos(θ) + sin(θ) + 4.
Step 3: Solve for f(θ)
Next, we integrate f'(θ) with respect to θ to obtain f(θ):
f(θ) = ∫(-cos(θ) + sin(θ) + 4) dθ
Using the integral rules, we get:
f(θ) = -sin(θ) - cos(θ) + 4θ + D
Step 4: Finding the constant of integration (D)
Substituting θ = 0 into f(θ) and equating it to the given f(0) = 5, we have:
-0 - 1 + 0 + D = 5
D = 6
So, the correct expression for f(θ) is:
f(θ) = -sin(θ) - cos(θ) + 4θ + 6.
Therefore, your expression for f(θ) was incorrect, and the correct expression is f(θ) = -sin(θ) - cos(θ) + 4θ + 6.
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