a) Find an equation of a line that is tangent to y=2sinx and whose slope is a maximum value.
b) Is there more than one solution? Explain.
remember that the slope of a function is the first derivative of that function.
So for the slope to have a maximum, the derivative of slope must be zero
But the derivative of the slope is the second derivative of the given function, so ...
y = 2sinx
y ' = 2 cosx
y'' = -2sinx
-2sinx = 0
sinx = 0
x = 0, π, 2π, 3π, 4π etc , for each of those y = 0
points where the slopes are a max are (0,0), (2π,0), (4π,0) etc.
Subbing each of those into the slope of 2cosx gives
a slope of 2(1) or 2
(notice that the slope at the points (π,0), (3π,0) etc would give us a slope of -2, thus it would be a minimum slope at those points)
The first line would be
y = 2x + b, with (0,0) on it
0 = 0 + b
b = 0
first tangent is y = x
second tangent:
y = 2x + b, with (2π,0) on it
0 = 2(2π) + b
b = -4π
second tangent equation:
y = 2x - 4π
etc.
you mean y=2x for the first line...because I did something similar however I took the limit, since cos x can not have y values larger than 1 or -1, i took 2cosx=1 i get zero and using both methods:Y-Y1=m(X-X1) and y=mx+b i get y=2x. i checked it using a graphical calculator the same answer, y=2x
a) Well, finding the equation of a line that is tangent to y=2sinx and has a maximum slope can be quite challenging! It's like trying to catch a unicorn at a speed-dating event. But fear not, I shall come to your rescue!
To find the maximum slope, we need to determine the point on the curve where the slope is at its peak. In other words, we're looking for the peak of the roller coaster ride that is this curve. Hang tight!
Let's differentiate y=2sinx with respect to x, using the chain rule. The derivative is d/dx(y) = 2cosx. Now, for the slope to be maximal, we need the derivative to be equal to zero, making it a peak! So, we solve 2cosx = 0 for x.
Now, cosx = 0 for x = (2n + 1)π/2, where n is an integer. These are the x-values where the slope is at its maximum.
To find the corresponding y-values, we substitute these x-values back into the equation y=2sinx. For example, when x = π/2, y = 2sin(π/2) = 2. Voila! Now you have a point where the tangent line can touch the curve.
Finally, we use the point-slope form of a line -- y - y1 = m(x - x1) -- where (x1, y1) is our point and m is the maximum slope we calculated earlier. Plug in the values, and you'll have your equation of the tangent line at its maximum slope!
b) As for multiple solutions, think of it this way: imagine you have a contest to find the luckiest leprechaun in the world. Now, if you have two or more leprechauns who are equally lucky, it's natural to have multiple winners, right?
Similarly, if there are multiple points on the sine curve where the slope is maximum, we'll end up with more than one line tangent to the curve. So, yes, there can be more than one solution to this problem, depending on how many lucky spots we find on the curve!
a) To find an equation of a line that is tangent to y = 2sin(x) and has a maximum slope, we need to determine the maximum slope of the function y = 2sin(x).
The slope of a function can be found by taking the derivative of the function with respect to x.
Let's find the derivative of y = 2sin(x):
dy/dx = d/dx(2sin(x))
Using the chain rule, we have:
dy/dx = 2 * cos(x) * d/dx(x)
= 2 * cos(x)
To find the maximum slope, we need to find the maximum value of cos(x).
The maximum value of cos(x) is 1, which occurs when x = 0.
Therefore, the maximum slope of y = 2sin(x) is 2 * 1 = 2.
To find the equation of the line tangent to y = 2sin(x) with a slope of 2, we can use the point-slope form of a line.
Since the line is tangent to the curve at x = 0, we have a point (0, 2sin(0)) = (0, 0) on the line.
Using the point-slope form, we can write the equation of the line as y - 0 = 2(x - 0), which simplifies to y = 2x.
Therefore, the equation of the line that is tangent to y = 2sin(x) with a maximum slope is y = 2x.
b) No, there is only one solution. The equation y = 2x represents a straight line, and there can only be one line tangent to a given point on a curve with a specific slope. In this case, the slope of the line is the maximum slope of the function y = 2sin(x), and it is unique.
a) To find an equation of a line that is tangent to the curve y = 2sin(x) and has the maximum slope, we can start by finding the derivative of the given function. The derivative will give us the slope of the tangent line at any point on the curve.
The derivative of y = 2sin(x) can be found using the chain rule, which states that if y = f(g(x)), then dy/dx = f'(g(x)) * g'(x). In this case, g(x) = x and f(x) = 2sin(x).
To find the derivative, we differentiate f(x) = 2sin(x) with respect to x:
f'(x) = d/dx(2sin(x))
= 2 * cos(x)
Now, we have the derivative of y = 2sin(x) as f'(x) = 2cos(x).
To find the maximum slope, we set the derivative equal to zero and solve for x:
2cos(x) = 0
Since cos(x) is equal to zero when x = π/2 + kπ and x = 3π/2 + kπ, where k is an integer, we have two potential maximum points.
To find the maximum slope, we can evaluate the derivative at these two points:
f'(π/2) = 2cos(π/2) = 0
f'(3π/2) = 2cos(3π/2) = 0
Both of these points have a slope of zero, which means they are not the maximum slope. However, let's evaluate the slope at a point near these values to determine the maximum slope.
Let's consider the point x = π/4:
f'(π/4) = 2cos(π/4) = √2
This value of √2 is the maximum slope.
Now that we have the maximum slope, we need to find the equation of the tangent line that has this slope and passes through a point on the curve y = 2sin(x).
To find the equation of the tangent line, we need a point (x1, y1) on the curve. Let's consider x = π/4:
y = 2sin(π/4) = 2/√2 = √2
So, we have the point (π/4, √2).
The equation of the line can be written as y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line.
Plugging in the values, we have:
y - √2 = √2(x - π/4)
Simplifying it, we get:
y = √2x - π/2
Therefore, the equation of the tangent line with the maximum slope is y = √2x - π/2.
b) No, there is only one solution because we found the maximum slope by evaluating the derivative of y = 2sin(x) and setting it equal to zero. This gave us two potential maximum points, but upon further evaluation, we found that both of them have a slope of zero. The maximum slope of √2 was obtained at the point x = π/4, and we derived the equation of the tangent line passing through this point and having the maximum slope. Therefore, there is only one solution.