Roman's Responses in WyzAnt Answers

This one is actually a classic problem for cross products.

<1, 0, -1> × <-2, 2, 0>

      |  i     j    k  |

=    |  1    0   -1 |

      | -2    2    0 |

= <0*0 - (-1)*2, (-1)*(-2) - 1*0, 1*2 - 0*(-2)>

= <2, 2, 2>

Area = ||<2,2,2>|| = √(2² + 2² + 2²) = 2√3.

By the definition of derivatives, you get the following

lim_Δx→0 [ sin( x + Δx ) - sin x ] / Δx = d/dx sin x = cos x, assuming that the angle x is measured in radians.

If you want to derive it, you will need to use the angle sum formula which is

sin(a+b) = sin a cos b + cos a sin b.

You get

lim_Δx→0 [ sin( x + Δx ) - sin x ] / Δx

= lim_Δx→0 [ sin x cos Δx + sin Δx cos x - sin x ] / Δx

= lim_Δx→0 (sin x)(cos Δ - 1) / Δx + lim _Δx→0 (cos x)(sin Δx) / Δx

= (sin x) lim_Δx→0 (cos Δx - 1) / Δx + (cos x) lim _Δx→0 (sin Δx) / Δx

Now all we need is to calculate the two limits

lim_Δx→0 (sin Δx) / Δx can be computed by drawing a picture. I won't draw it here, but from the correct picture you get that

sin Δx < Δx < tan Δx for 0 < Δx < Π/2

and since sin Δx, Δx, and tan Δx are odd functions,

sin Δx > Δx > tan Δx for -Π/2 < Δx < 0.

Dividing either inequality by Δx and taking reciprocal gives.

1 > (sin Δx) / Δx > cos Δx for -Π/2 < Δx < Π/2 and Δx ≠ 0.

Since lim_Δx→0 cos Δx = 1, the squeezing theorem implies that

lim_Δx→0 (sin Δx) / Δx = 1.

We can use this limit to evaluate the other limit as follows.

lim_Δx→0 (cos Δx - 1) / Δx

= lim_Δx→0 (cos Δx - 1)(cos Δx + 1) / [Δx (cos Δx + 1)]

= lim_Δx→0 (cos² Δx - 1) / [Δx (cos Δx + 1)]

= lim_Δx→0 (-sin² Δx) / [Δx (cos Δx + 1)]

= -(lim_Δx→0 (sin Δx) / Δx)(lim_Δx→0 (sin Δx) / (cos Δx + 1))

= 1*(sin 0) / (cos 0 + 1)

= 0 / 2

= 0

Plugging these limits in gives

lim_Δx→0 [ sin( x + Δx ) - sin x ] / Δx = 0 sin x + 1 cos x = cos x.

lim_θ→Π/2 (cot θ - cos θ) / cos³θ

= lim_θ→Π/2 (csc θ - 1) cos θ / cos³ θ

= lim_θ→Π/2 (csc θ - 1) / cos² θ

= lim_θ→Π/2 (1 - sin θ) csc θ / (1 - sin² θ)

= lim_θ→Π/2 (1 - sin θ) csc θ / [(1 - sin θ)(1 + sin θ)]

= lim_θ→Π/2 csc θ / (1 + sin θ)

= csc(Π/2) / [1 + sin(Π/2)]

= 1 / (1+1)

= 1/2 ← Answer

Start by making a table like so.

             B:Yes         B:No

A:Yes       W              X

A:No         Y               Z

We know that 15% have disease A, so 1: W+X = 15%.

We know that 20% have disease B, so 2: W+Y = 20%.

We know that 25% have at least one of the two diseases, so 3: W+X+Y = 25%.

The four entries in the table total the entire population, so 4: W+X+Y+Z = 100%

Subtracting equation 2 from equation 3 gives us X = 5%.

Subtracting equation 1 from equation 3 gives us Y = 10%.

Plugging these in to equation 3 gives W + 5% + 10%= 25%, so W = 10%.

Finally, subtracting equation 3 from equation 4 gives us Z = 75%.

We can now fill in the table:

               B:Yes         B:No

A:Yes        10%          5%

A:No         10%          75%

You should now be able to answer both questions.

First of all, note that it will be a parabola, and that (2,-2) is it's focus and x - y = 0 is it's directrix.

Let's work with the squares of the distances instead since then there won't be any radicals.

Given a point (x,y) on this parabola, we have the squared distance from the focus as

D²_focus = (x - 2)² + (y + 2)².

To get the squared distance from the directrix, D²_directricx , we we can use vectors.

Pick a vector perpendicular to the directrix.

Recall that vector < a , b > is perpendicular to line ax + by = c

For example, v = < 1 , -1 > is perpendicular to our directrix x - y = 0.

Let u = < x , y >.

Then the desired distance is D_directricx = u · v / ||v||.

Thus D²_directricx = (<x , y> · <1 , -1> / || <1 , -1> ||)²

= (x - y)² / 2

The locus you are looking for is the same as D²_focus = D²_directrix which is

2[(x - 2)² + (y + 2)²] = (x - y)²

2x² - 8x + 8 + 2y² + 8y + 8 = x² - 2xy + y²

x² + 2xy + y² - 8x + 8y + 16 = 0