Well, well, well, looks like we've got a curve that's causing some curveballs, huh? Don't worry, I'll do my best to help you out.
a. To find dy/dx in terms of y, we can differentiate both sides of the equation with respect to x. Remember to use the chain rule if necessary. To be honest, though, your answer of 1/(y - sin(y)) doesn't quite seem right. Instead, let's give it another try.
Differentiating y + cosy = x + 1 with respect to x, we get:
dy/dx - sin(y) * (dy/dx) = 1.
Now, we can rearrange and solve for dy/dx:
dy/dx * (1 - sin(y)) = 1.
dy/dx = 1 / (1 - sin(y)).
So, the correct answer is dy/dx = 1 / (1 - sin(y)).
b. Now, let's tackle those vertical tangents. Remember, vertical tangents occur when the derivative is undefined or approaches infinity. In other words, we want to find values of y that make the denominator of dy/dx = 0.
To find those values, we solve the equation 1 - sin(y) = 0:
sin(y) = 1.
Now, in the interval 0 ≤ y ≤ 2π, you'd expect sin(y) to never be 1 because, well, we're dealing with real numbers here. However, sin(y) equals 1 when y = π/2. So, we have a vertical tangent at y = π/2.
c. Finally, it's time to find d^2y/dx^2 in terms of y. This means we need to differentiate dy/dx with respect to x again. Brace yourself!
Differentiating dy/dx = 1 / (1 - sin(y)) with respect to x, we get:
d^2y/dx^2 = 0.
Surprise! The second derivative turns out to be 0 no matter what value of y you plug in. So, d^2y/dx^2 = 0.
d. Now, for the grand finale, let's sketch a graph of the curve using a table of values. Choose some values for y within the range 0 ≤ y ≤ 2π and plug them into the equation y + cosy = x + 1 to find the corresponding x values. Then plot those points and connect the dots with a smooth curve.
Here's a little data for you:
y = 0 -> x = 0, (0, 0)
y = π/4 -> x ≈ -0.366, (-0.366, π/4)
y = π/2 -> x ≈ 0.841, (0.841, π/2)
y = 3π/4 -> x ≈ 1.79, (1.79, 3π/4)
y = π -> x ≈ 3.56, (3.56, π)
y = 5π/4 -> x ≈ 5.74, (5.74, 5π/4)
y = 3π/2 -> x ≈ 6.80, (6.80, 3π/2)
y = 7π/4 -> x ≈ 7.71, (7.71, 7π/4)
y = 2π -> x ≈ 8, (8, 2π)
Now, grab your trusty ruler and draw a smooth curve passing through those points. Voila! You've got yourself a sketch of the curve.
I hope this helps, and remember, if calculus ever gets too serious, just add a little humor to solve for funny!