find the point (or points) where the ellipse 9x^2+4y^2=36 has maximum curvature. don't even know where to start with this one.
just a note i now know how to do it but the method to do it will literally take up like 4 pages. if i solve for y=f(x) and use the equation for the curve, k. please tell me im missing something that will make this simpler.
You want dy/dx to be a maximum
So the 2nd derivative has to be zero
instead of solving for y, differentiatte implicitly
18x + 8y dy/dx = 0
dy/dx = =18x/8y = -9x/4y
second derivative = (4y(-9) - (-9x)(4dy/dx) )/16y^2
= (-36y + 36x dy/dx)/(16y^2)
= 0 for a max/min of dy/dx
36x dy/x = 36y
x dy/dx = y
dy/dx = y/x
but dy/dx = -9x/4y
continue please
ok got it. thank you. i also solve for y and plug it in right? so i can solve for x.
reiny is wrong you cannot just use the second derivative to find the maximum curvature.
To find the point(s) of maximum curvature for the given ellipse, we can follow these steps:
Step 1: Rewrite the given ellipse equation in the standard form.
Step 2: Calculate the curvature equation for the curve.
Step 3: Take the derivative of the curvature equation and find the critical points.
Step 4: Determine whether the critical points correspond to a relative maximum.
Let's go through each step in detail.
Step 1: Rewrite the ellipse equation in standard form.
The given ellipse equation is 9x^2 + 4y^2 = 36. To rewrite it in standard form, we need to isolate the x^2 and y^2 terms and divide both sides by 36 to make the equation equal to 1. Dividing throughout by 36 gives:
x^2/4 + y^2/9 = 1
Step 2: Calculate the curvature equation for the curve.
The curvature equation for an ellipse can be represented as:
K = |(x'^2 * y'' - y'^2 * x'') / (x'^2 + y'^2)^(3/2)|
where x' and y' represent the first derivatives of x and y with respect to the parameter (usually t or θ), and x'' and y'' represent the second derivatives.
In our case, the coordinate representation of the ellipse x(t) = 2cos(t) and y(t) = 3sin(t) allows us to use the parameter t. Therefore, we need to find x' = dx/dt and y' = dy/dt first.
Taking the first derivatives gives:
x' = -2sin(t)
y' = 3cos(t)
Now, differentiate x' and y' to find x'' and y''.
x'' = -2cos(t)
y'' = -3sin(t)
Step 3: Take the derivative of the curvature equation and find the critical points.
Substituting the derivatives into the curvature equation, we get:
K = |((-2sin(t))^2 * (-3sin(t)) - (3cos(t))^2 * (-2cos(t))) / ((-2sin(t))^2 + (3cos(t))^2)^(3/2)|
Simplifying the equation further, we have:
K = |(12sin(t)^4 - 4cos(t)^4) / (4sin(t)^2 + 9cos(t)^2)^(3/2)|
To find the critical points, we need to find where the derivative of K with respect to t is equal to zero. You can solve this equation by taking the derivative of K and setting it equal to zero:
dK/dt = 0
Step 4: Determine whether the critical points correspond to a relative maximum.
After solving the equation, you will find values of t that correspond to critical points. Evaluate the second derivative of K at these critical points to determine whether they correspond to a relative maximum or not:
d^2K/dt^2
If d^2K/dt^2 > 0, then the critical point corresponds to a relative maximum.
By following these steps, you should be able to find the point(s) on the ellipse where maximum curvature occurs.