Find the equation of the curve that passes through the point (x, y) = (0, 0) and has an arc length on the interval 0⩽x⩽π/4 given by the integral from 0 to π/4 of√(1+cos^2x) dx.
a) y= sin(x) ------> My answer. Can you check for me, pleaseeee?
b) y= cos(x)
c) y= cos^-1 (x)
d) y= tan(x)
since arc length is ∫√(1+(y')^2) dx
y' = cosx
so, y = sinx
I agree.
of course, y = -sinx would also work, but that was not one of the choices ...
To find the equation of the curve that passes through the point (0, 0) and has a specified arc length, we need to evaluate the integral given.
The integral given is ∫[0 to π/4] √(1+cos^2(x)) dx.
To evaluate this integral, let's go step by step. First, note that the integrand contains a trigonometric term (cos^2(x)), so we need to simplify it before integrating.
Using the trigonometric identity cos^2(x) = 1 - sin^2(x), we can rewrite the integrand as follows:
∫[0 to π/4] √(1+cos^2(x)) dx = ∫[0 to π/4] √(1+(1- sin^2(x))) dx
Simplifying further, we have:
∫[0 to π/4] √(2 - sin^2(x)) dx
Now, let's try to tackle this integral. We can use a substitution to simplify it further.
Let's substitute u = sin(x), which implies du = cos(x) dx.
When x = 0, u = sin(0) = 0, and when x = π/4, u = sin(π/4) = 1/√2.
The integral becomes:
∫[0 to π/4] √(2 - sin^2(x)) dx = ∫[0 to 1/√2] √(2 - u^2) (1/cos(x)) du
Note that we substitute cos(x) dx = du and use the identity cos(x) = 1/sqrt(1 - sin^2(x)).
Now, we can integrate with respect to u:
∫[0 to 1/√2] √(2 - u^2) (1/cos(x)) du = ∫[0 to 1/√2] √(2 - u^2) du
To evaluate this integral, we recognize that the integrand represents a semi-circle of radius √2.
The equation of a circle with the center at the origin (0, 0) and a radius of √2 is given by x^2 + y^2 = 2.
The equation of the curve passing through (0, 0) and having the specified arc length is x^2 + y^2 = 2 (equation of a circle) or equivalently, y = ± √(2 - x^2).
Therefore, the correct answer is not listed among the choices. The equation of the curve is y = ± √(2 - x^2).