Hi! Please help :-)
Find an equation for the circle of the curvature of the curve
r(t) = ti + (sint)j at the point (pi/2, 1)
(The curve parametrizes the graph of y = sinx in the xy-plane)
To find the equation for the circle of curvature of a given curve, we need to find the center and radius of the circle.
Let's start by finding the radius of curvature (ρ). The formula for the radius of curvature is:
ρ = (1 + (dy/dx)^2)^(3/2) / (|d^2y/dx^2|)
In this case, the curve is parametrized as r(t) = ti + (sin(t))j.
Let's find the first derivative of y with respect to x (dy/dx):
dy/dx = (dy/dt) / (dx/dt)
dy/dt = (dr(t)/dt).j
dr(t)/dt = i + (cos(t)).j
dx/dt = 1
Therefore, dy/dx = (i + (cos(t)).j) / 1
= i + (cos(t)).j
Now, let's find the second derivative of y with respect to x (d^2y/dx^2):
d^2y/dx^2 = (d^2y/dt^2) / (dx/dt)
d^2y/dt^2 = (d(dr(t)/dt).j)/(dx/dt)
d(dr(t)/dt)/dt = (d(i + (cos(t)).j)/dt)
= - (sin(t)).i
d^2y/dx^2 = (- (sin(t)).i) / 1
= - (sin(t)).i
Now, let's find the value of t at the point (π/2, 1) to substitute into the above formulas:
For x = π/2, t = π/2
Therefore, at the point (π/2, 1), t = π/2.
Substituting t = π/2 into the formulas for dy/dx and d^2y/dx^2, we get:
dy/dx = i + (cos(π/2)).j
= i + 0.j
= i
d^2y/dx^2 = - (sin(π/2)).i
= - i
Now, let's find the radius of curvature (ρ) using the formula:
ρ = (1 + (dy/dx)^2)^(3/2) / (|d^2y/dx^2|)
= (1 + (i)^2)^(3/2) / |-i|
Since (i)^2 = -1, we have:
ρ = (1 + (-1))^(3/2) / |-i|
= 0^(3/2) / 1
= 0 / 1
= 0
Therefore, the radius of curvature at the point (π/2, 1) is 0.
Since the radius of curvature is 0, this means that the equation for the circle of curvature of the curve at this point is simply the point itself. In this case, the equation for the circle of curvature is:
(x - π/2)^2 + (y - 1)^2 = 0
To find the equation for the circle of curvature of a given curve, we need to find the radius and center of the circle at a specific point on the curve. In this case, we need to find the equation for the circle of curvature at the point (pi/2, 1) on the curve y = sin(x).
To find the radius of curvature, we can use the formula:
ρ = (1 + (dy/dx)^2)^(3/2) / |d^2y/dx^2|
where dy/dx represents the derivative of y with respect to x, and d^2y/dx^2 represents the second derivative of y with respect to x.
Let's find the values of dy/dx and d^2y/dx^2 for the given curve.
Given curve: y = sin(x)
Taking the derivative of y with respect to x:
dy/dx = cos(x)
Taking the derivative of dy/dx with respect to x:
d^2y/dx^2 = -sin(x)
Now, let's find the values of dy/dx and d^2y/dx^2 at the point (pi/2, 1).
Substituting x = pi/2 into dy/dx:
dy/dx = cos(pi/2) = 0
Substituting x = pi/2 into d^2y/dx^2:
d^2y/dx^2 = -sin(pi/2) = -1
Now, we can find the radius of curvature using the formula:
ρ = (1 + (dy/dx)^2)^(3/2) / |d^2y/dx^2|
Plugging in the values we found:
ρ = (1 + (0)^2)^(3/2) / |-1|
ρ = 1 / 1
ρ = 1
The radius of curvature at the point (pi/2, 1) is 1.
To find the center of the circle of curvature, we need the normal vector to the curve at the point (pi/2, 1). The normal vector can be obtained by taking the negative reciprocal of dy/dx.
At the point (pi/2, 1), dy/dx = 0. The negative reciprocal of 0 is undefined. However, we know that the normal vector at this point will be a vertical vector, as the curve is orthogonal to the x-axis.
Therefore, the center of the circle of curvature is located on the straight line passing through the point (pi/2, 1) and is parallel to the x-axis.
In conclusion, the equation for the circle of curvature at the point (pi/2, 1) on the curve y = sin(x) is x = pi/2.