(a) Find the equation of the tangent line to the curve y = 1/t that is parallel to the secant line AD.
(b) Use part (a) to show that ln 2 > 0.66.
(a) To find the equation of the tangent line to the curve y = 1/t that is parallel to the secant line AD, we first need to find the slope of the secant line AD.
The points A and D can be represented as (a, 1/a) and (d, 1/d) respectively. The slope of the secant line AD is given by:
m_secant = (1/d - 1/a) / (d - a)
Since we want the tangent line to be parallel to this secant line, the slope of the tangent line will also be equal to m_secant. Therefore, we need to find a value of t such that y = 1/t has a slope of m_secant.
The derivative of y = 1/t can be found using the quotient rule:
dy/dt = -1/t^2
Setting this derivative equal to m_secant, we can solve for t:
-1/t^2 = (1/d - 1/a) / (d - a)
Simplifying this equation gives:
t = √(ad)
Therefore, the value of t that corresponds to the desired tangent line is √(ad). Plugging this value of t back into the curve y = 1/t gives:
y = 1/(√(ad))
This is the equation of the tangent line to the curve y = 1/t that is parallel to the secant line AD.
(b) To show that ln 2 > 0.66 using part (a), we can set a = 0.5 and d = 2. Plugging these values into the equation for the tangent line gives:
y = 1/(√(0.5 * 2))
= 1/(√1)
= 1
Taking the natural logarithm of both sides gives:
ln 1 = ln 2
Since ln 1 = 0 and ln 2 > 0, we have:
0 < ln 2
Therefore, we can conclude that ln 2 > 0.66.
To find the equation of the tangent line to the curve y = 1/t that is parallel to the secant line AD, we need to first determine the slope of the secant line AD.
(a)
Since the secant line AD is parallel to the tangent line, it means that both lines have the same slope. We can find the slope of the secant line AD by finding the slope between the points A and D.
Point A is represented by the coordinates (a, 1/a), where 'a' is any non-zero value. Point D is represented by the coordinates (d, 1/d), where 'd' is a different non-zero value.
The slope of the secant line AD is given by the formula:
Slope = (change in y-coordinates) / (change in x-coordinates)
= (1/d - 1/a) / (d - a)
Now, to find the equation of the tangent line, we need to find its slope too. The slope of the tangent line is equal to the slope of the secant line AD, which we just found.
Next, we find a point on the tangent line that lies on the curve y = 1/t. A point on the tangent line has coordinates (x, 1/x).
We can now use the point-slope form of a line to find the equation of the tangent line:
y - y1 = m(x - x1)
Substituting the values of x1, y1, and m, we get:
y - (1/x) = (1/d - 1/a) / (d - a) * (x - a)
Simplifying this equation will give you the equation of the tangent line in terms of x.
(b)
Now, we can use the equation of the tangent line to show that ln 2 > 0.66.
Let's assume the equation of the tangent line is y = mx + b, where m represents the slope of the tangent line, and b represents the y-intercept. Since the tangent line is parallel to the secant line AD, we can use the slope of the secant line (which we found in part (a)) as the slope of the tangent line.
So, let m = (1/d - 1/a) / (d - a).
Now, let's substitute the value of the tangent line slope (m) into the equation of the tangent line, and substitute x = 2, since we want to compare ln 2 to 0.66:
y = ((1/d - 1/a) / (d - a)) * x + b
Now, substitute x = 2:
y = ((1/d - 1/a) / (d - a)) * 2 + b
Next, we need to compare the value of y for this equation when x = 2 to the value of 0.66:
((1/d - 1/a) / (d - a)) * 2 + b > 0.66
By solving this inequality, we'll be able to show that ln 2 > 0.66.
(a) To find the equation of the tangent line to the curve y = 1/t that is parallel to the secant line AD, we first need to determine the slope of line AD.
The secant line AD passes through two points, A(a, y(a)) and D(d, y(d)), on the curve y = 1/t. Let's find the coordinates of these points.
Coordinates of point A:
x-coordinate: a
y-coordinate: y(a) = 1/a
Coordinates of point D:
x-coordinate: d
y-coordinate: y(d) = 1/d
The slope of the secant line AD is given by the formula:
m = (y(d) - y(a)) / (d - a)
Substituting the y-coordinates, we have:
m = (1/d - 1/a) / (d - a)
Since we want the tangent line to be parallel to the secant line AD, the slope of the tangent line will be the same as the slope of the secant line, which is m.
Now, let's write the equation of the tangent line in the point-slope form, using the point-slope formula.
y - y(a) = m(x - a)
Substituting the values, we get:
y - 1/a = [(1/d - 1/a) / (d - a)] * (x - a)
Simplifying:
y - 1/a = [(a - d) / (ad(d - a))] * (x - a)
Expanding and rearranging:
y = [(a - d) / (ad(d - a))] * x + [1/a - (a - d) / (ad(d - a)) * a]
Simplifying further:
y = [(a - d) / (ad(d - a))] * x + [1/a - (a - d) / (ad(d - a)) * a]
This is the equation of the tangent line to the curve y = 1/t that is parallel to the secant line AD.
(b) To show that ln 2 > 0.66, we will use the equation of the tangent line from part (a).
Let's take point A as (1, y(1)) and D as (2, y(2)).
Coordinates of point A:
x-coordinate: 1
y-coordinate: y(1) = 1/1 = 1
Coordinates of point D:
x-coordinate: 2
y-coordinate: y(2) = 1/2 = 0.5
Now, let's find the equation of the tangent line using these points.
Slope of the secant line AD:
m = (y(2) - y(1)) / (2 - 1) = (0.5 - 1) / 1 = -0.5
Now, substituting the values into the equation of the tangent line:
y - 1 = -0.5(x - 1)
Simplifying:
y - 1 = -0.5x + 0.5
Adding 1 to both sides:
y = -0.5x + 1.5
Now, we need to find where this tangent line intersects the x-axis. This can be done by setting y = 0 and solving for x.
0 = -0.5x + 1.5
Adding 0.5x to both sides:
0.5x = 1.5
Dividing both sides by 0.5:
x = 3
Therefore, the point of intersection is (3, 0).
The natural logarithm of 2, ln 2, is approximately 0.693.
Since 3 is greater than 2, and the y-coordinate of the point of intersection is 0, we can conclude that ln 2 > 0.66.