Notes from Class 1209


Calculus I Quiz 1209

Sketch the curve y = sqrt(x) for 0 < x < 16.

Construct a rectangle by sketching a vertical line from the point (.5,0) to the curve, then right to the vertical line x = 16.  What is the area of this rectangle?

At x = .5 we have sqrt(x) = sqrt(.5) =.71 approx.

The rectangle has altitude .71 and extends from x = .5 to x = 4. 

Area is therefore .71 ( 3.5) = 2.5 approx.

Repeat for the point (1, 0).

Dimensions 1 x 3, area 3.

Repeat for the point (2, 0).

Dimensions sqrt(2) x 2, area 2 sqrt(2) = 2.8 approx..

Repeat for the point (3, 0).

Dimensions sqrt(3) x 1, area sqrt(3) = 1.7 approx..

These dimensions and areas are summarized in the picture below.

  

Repeat for the point (a, 0).  What value of a maximizes the area of the rectangle?

As shown in the figure below we find that:

It follows that the area is maximized for a = 4/3.

The functions sinh(x) and cosh(x) are defined as follows:

Show that the derivative of sinh(x) is equal to cosh(x), while the derivative of cosh(x) is equal to the derivative of sinh(x).

[ sinh(x) ] ' = [ (e^x - e^-x) / 2 ] ' = 1/2 (e^x - e^-x) ' = 1/2 (e^x + e^-x) = (e^x + e^-x) / 2 = cosh x.

[ cosh(x) ] ' = [ (e^x + e^-x) / 2 ] ' = 1/2 (e^x + e^-x) ' = 1/2 (e^x - e^-x) = (e^x - e^-x) / 2 = sinh x.

If tanh(x) = sinh(x) / cosh(x) then what is the derivative of tanh(x)?

We easily find the derivative using the quotient rule.  The derivative is found in the figure below.

We note that the identity cosh^2(x) - sinh^2(x) = 1, which is somewhat analogous to the identity cos^2(x) + sin^2(x) = 1, is easily proven from the definitions of the cosh and sinh functions.

There are two forms in which the derivative of tanh(x) can be expressed.  One is 1 / cosh^2(x), the other is 1 - tanh^2(x).