(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.
(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.
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