Of the infinitely many lines that are tangent to the curve y = −6 sin x and pass through the origin, there is one that has the largest slope. Use Newton's method to find the slope of that line correct to six decimal places.
let the point of contact be (x,y)
or (x, -6sinx)
Then for a tangent,
slope = m = (-6sinx - 0)/(x-0) = -6sinx/x
m = (-6sinx - 0)/(x-0) = -6sinx/x
dm/dx = (x(-6cosx - (-6sinx))/x^2 , by quotient rule
= 0 for a max/min
-6xcosx + 6sinx = 0
xcosx - sinx = 0
xcosx = sinx
x = sinx/cosx
tanx = x
let y = tanx - x
y' = sec^2 x - 1
Newton's Method"
x2 = x1 - y/y'
= x1 - (tan x1 - x1)/(sec^2 x1 - 1)
according to my sketch, x is between π and 2π, so how about
x1 = 4.5
x2 = 4.4936...
x1 = 4.4936
x2 = 4.4934096
x1 = 4.4934096
x2 = 4.4934095
how about that???
so m = -6sinx/x = 1.303402
and the tangent equation would be y = 1.303402x
check with Wolfram:
http://www.wolframalpha.com/input/?i=plot+y+%3D+-6sinx+%2C+y+%3D+1.3034x+%2C+0+%3C+x+%3C+10
Now , how good is that?
btw:
http://www.wolframalpha.com/input/?i=tanx+%3D+x
look at the 3rd solution.
we came up with that solution, had our guess been close to one of the other solutions, our Newton's Method would have zeroed in on the one closest to our guess.
To find the slope of the tangent line to the curve y = -6 sin(x) that passes through the origin, we can start by finding the derivative of the curve.
The derivative of y = -6 sin(x) can be found by applying the chain rule. Recall that the derivative of sin(x) is cos(x) and the derivative of -6x is -6. Thus, the derivative of y = -6 sin(x) is dy/dx = -6 cos(x).
Now, let's use Newton's method to find the value of x where the slope of the tangent line is at its maximum. Newton's method is an iterative technique for finding the roots of a function.
1. Start with an initial guess for x, denoted as x0. Since we want to find the tangent line that passes through the origin, we can guess x0 = 0.
2. Compute the function value and the derivative at the current guess: f(x0) = -6 cos(x0) and f'(x0) = 6 sin(x0).
3. Update the guess for x using the formula:
x1 = x0 - (f(x0) / f'(x0))
4. Repeat steps 2 and 3 until convergence is achieved.
Let's perform the iterations using Newton's method to find the maximum slope.
Iteration 1:
x0 = 0
f(x0) = -6 cos(0) = -6
f'(x0) = 6 sin(0) = 0
x1 = 0 - (-6 / 0) = undefined
Since the slope is undefined in iteration 1, let's try a new initial guess.
Iteration 2:
x0 = π/2 (Any value between 0 and π/2 will work.)
f(x0) = -6 cos(π/2) = 0
f'(x0) = 6 sin(π/2) = 6
x1 = π/2 - (0 / 6) = π/2
Iteration 3:
x0 = π/2
f(x0) = -6 cos(π/2) = 0
f'(x0) = 6 sin(π/2) = 6
x1 = π/2 - (0 / 6) = π/2
Since the guess for x did not change in iteration 3, we can conclude that the maximum slope of the tangent line is 6.
Therefore, the slope of the tangent line that has the largest slope and passes through the origin is 6.