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.
Naturally, the tangent with the largest slope will be the one closest to the origin.
So, we want
-6cosx = -6sinx/x
x = tanx
solve that using Newton's method, and you have x = 4.93
so, your line is
y = -6cos(4.93)(x-4.93)+6sin(4.93)
or
y-5.98 = 0.52(x-4.93)
See the graphs at
http://www.wolframalpha.com/input/?i=plot+y+%3D+-6sin%28x%29%2C+y%3D0.52%28x-4.93%29%2B5.98+where+0%3C%3Dx%3C%3D2pi
if you tighten your accuracy, you should get a better tangent.
Wasn't your tangent line supposed to pass through the origin?
Reiny, yes it does pass through the origin.
I sure flubbed that one. Your later solution was spot on, Reiny.
To find the line with the largest slope that is tangent to the curve and passes through the origin, we need to maximize the derivative of the curve. In this case, the curve is given by y = -6sin(x), so we need to find the maximum value of the derivative of this function.
First, let's find the derivative of y = -6sin(x). The derivative of sin(x) is cos(x), so the derivative of -6sin(x) is -6cos(x).
Now, we need to find the value of x where the derivative equals zero, i.e., where -6cos(x) = 0. From this equation, we can determine that cos(x) = 0. Solving for x, we find x = π/2 + nπ, where n is an integer.
To apply Newton's method, we need an initial guess for the value of x. Since we want to find the line with the largest slope, we can start with an initial guess of x = π/2.
Now, let's apply Newton's method:
1. Start with an initial guess for x, let's say x0 = π/2.
2. Use the formula x1 = x0 - f(x0)/f'(x0), where f(x) = -6cos(x), and f'(x) = 6sin(x).
In our case, x1 = x0 - (-6cos(x0))/(6sin(x0)).
Substitute x0 = π/2 into the formula:
x1 = π/2 - (-6cos(π/2))/(6sin(π/2)).
Simplifying further, we get:
x1 = π/2 + 6/6 = π/2 + 1 = 3π/2.
3. Repeat step 2 using x1 as the new guess for x, until you reach the desired level of accuracy.
We can continue this process iteratively until we achieve the desired level of accuracy, which is to six decimal places. However, in this case, since cos(3π/2) = 0, the derivative is already zero at x = 3π/2 and won't change further.
Now, let's find the slope of the tangent line at x = 3π/2. The slope of a line tangent to the curve y = -6sin(x) is given by the derivative of the function evaluated at that particular x value. So, the slope is given by -6cos(3π/2).
Evaluating this expression, we find:
-6cos(3π/2) = -6(0) = 0.
Therefore, the slope of the line tangent to the curve y = -6sin(x) at x = 3π/2 and passing through the origin is 0, correct to six decimal places.