Let f(x) = e^x and g(x) = sin(lnx). Find equations for two distinct parallel lines l1 and l2 where; l1 is tangent to f at x = 0 and l2 is tangent to g.
f'(x) = e^x
f'(0) = 1
g'(x) = cos(lnx)/x
cos(lnx)/x=1
cos(lnx)=x
x = 0.0089,0.2745
So, you want lines with slope=1 passing through
(0,1),(0.0089,0.9933),(0.2745,-0.9610)
y-1 = x
y-0.9933 = x-0.0089
y+0.9610 = x-0.2745
See the graphs at
http://www.wolframalpha.com/input/?i=plot+y%3Dsin(lnx),y%3De%5Ex,y%3Dx%2B0.9910,+y%3Dx-1.2355+for+0.0080%3Cx%3C1
oops. there is also another solution at x=1 since cos(log1)/1 = cos(0)/1 = 1/1 = 1
To find the equations for the two distinct parallel lines, we first need to find the respective slopes of these lines. Let's start with line l1, which is tangent to f(x) = e^x at x = 0.
The slope of the tangent line l1 can be found by taking the derivative of the function f(x) = e^x with respect to x. Differentiating f(x) gives us:
f'(x) = d/dx(e^x).
The derivative of e^x is simply e^x. Therefore, the slope of l1 is given by:
m1 = f'(0) = e^0 = 1.
So, line l1 has a slope of 1.
Now, let's move on to finding the slope of line l2, which is tangent to g(x) = sin(lnx).
The slope of the tangent line l2 can be found by taking the derivative of the function g(x) = sin(lnx) with respect to x. This involves chain rule as we differentiate the composite function.
g'(x) = d/dx(sin(lnx)) = cos(lnx) * (1/x).
To find where l2 is tangent to g, we need to find the x-coordinate of the tangent point. In this case, the x-coordinate is where g'(x) is equal to 0, indicating a horizontal tangent line.
0 = cos(lnx) * (1/x).
This implies that cos(lnx) = 0 or 1/x = 0.
Since cos(lnx) cannot be 0, we focus on the second condition: 1/x = 0 (which results in x = ∞).
Therefore, the slope of l2 is given by:
m2 = g'(∞) = cos(ln∞) * (1/∞) = 0.
So, line l2 has a slope of 0.
Now that we know the slopes of the two lines, we can write their equations in point-slope form.
For line l1, since it is tangent to f(x) = e^x at x = 0, we have the point (0, f(0)) = (0, 1), and the slope m1 = 1. Therefore, the equation of l1 is:
y - y1 = m1(x - x1),
y - 1 = 1(x - 0),
y - 1 = x.
For line l2, since it is tangent to g(x) = sin(lnx), we don't have a specific point, as it is tangent at infinity. However, since the slope m2 = 0, we can write the equation of l2 as:
y = g(∞) = sin(ln∞) = sin(∞).
The value of sin(∞) is not defined, but the equation of l2 can be written as:
y = k, where k is a constant.
Hence, the equations for the two distinct parallel lines are:
l1: y = x,
l2: y = k, where k is a constant.