Find an equation of a line that is tangent to y=2sinx and whose slope is a maximum
The slope is a max/min when the second derivative is zero
y' = 2cosx
y'' = -2sinx
-2sinx = 0
sinx = 0
x = 0, π, 2π , ..
when x=0, y' = 2cos0 = 2 , then slope = 2
when x = π, y' = 2cosπ = -2
let's use x=0, then y = 0
slope =2
y = 2x + b
at (0,0)
0 = 0+b
A tangent equation with a maximum slope is
y = 2x
Well, if we're looking for maximum slope, we might want to look at the derivative of the function y = 2sin(x). The derivative will give us the slope of the tangent line at any point. Let's find it:
The derivative of y = 2sin(x) with respect to x is found by using the chain rule:
dy/dx = 2cos(x)
Now, we want to find the maximum slope. Since the derivative is a cosine function and its maximum value is 1, the maximum slope is 2.
So, the equation of the line that is tangent to y = 2sin(x) and has a maximum slope of 2 is:
y = 2x + c
where c is the y-intercept.
To find the equation of a line that is tangent to the curve y = 2sin(x) and has a maximum slope, we need to find the point where the maximum slope occurs.
The slope of the curve y = 2sin(x) at any point (x, y) can be found using the derivative. Taking the derivative of y with respect to x gives:
dy/dx = 2cos(x)
To find the maximum slope, we need to find the point where the derivative dy/dx equals its maximum value. Since the cosine function has a maximum value of 1, the maximum slope occurs when cos(x) = 1. This happens when x = 0, or more generally, at any point x = 2πn, where n is an integer.
At x = 0, the y-coordinate of the curve y = 2sin(x) is y = 2sin(0) = 0.
So, the point where the maximum slope occurs is (0, 0). This will be the point where the line is tangent to the curve.
Now we have a point on the line and the slope of the line, which is the maximum slope of the curve at that point. We can use the point-slope form of a line to find the equation of the tangent line.
The point-slope form of a line is given by:
y - y₁ = m(x - x₁)
where (x₁, y₁) is a point on the line, m is the slope of the line.
Using the point (0, 0) and the slope m = dy/dx = 2cos(0) = 2, the equation of the line tangent to the curve y = 2sin(x) with maximum slope is:
y - 0 = 2(x - 0)
Simplifying this equation, we get:
y = 2x
Therefore, the equation of the line tangent to the curve y = 2sin(x) with a maximum slope is y = 2x.
To find an equation of a line that is tangent to the curve y = 2sin(x) and has a maximum slope, we need to use calculus.
Derivatives can help us determine the slope of a curve at any given point. Since we are looking for the maximum slope, we need to find where the derivative of the curve is equal to zero.
Step 1: Find the derivative of y = 2sin(x).
Let's find the derivative of y = 2sin(x) using the chain rule:
dy/dx = 2cos(x)
Step 2: Find the critical points.
To find where the derivative is equal to zero and get the maximum slope, we need to solve the equation 2cos(x) = 0.
Setting 2cos(x) = 0:
cos(x) = 0
The cosine function is equal to zero at x = π/2 and x = 3π/2. These values represent the critical points.
Step 3: Find the maximum slope.
To determine which critical point gives the maximum slope, we can evaluate the second derivative for each critical point.
The second derivative of y = 2sin(x) is:
d²y / dx² = -2sin(x)
Evaluating the second derivative at x = π/2:
d²y/dx² = -2sin(π/2) = -2
Evaluating the second derivative at x = 3π/2:
d²y/dx² = -2sin(3π/2) = 2
Since d²y/dx² = -2 at x = π/2, it represents a maximum slope.
Step 4: Find the equation of the tangent line.
Now, we can find the equation of the tangent line using the point-slope form.
We know that the tangent line passes through the point (π/2, 2sin(π/2)) = (π/2, 1).
Using the point-slope form:
y - y₁ = m(x - x₁)
y - 1 = -2(x - π/2)
Expanding the equation:
y - 1 = -2x + π
y = -2x + π + 1
Therefore, the equation of the line tangent to y = 2sin(x) with a maximum slope is y = -2x + π + 1.