TWO Calculus QTNs NEED answering! URGENT HELP REQUIRED! Please!?

Question

QUESTION 1

Two roads intersect at right angles, Caitlein and John are 100km away from the intersection. Caitlein is driving towards the road at 30km/h and John at 40 km/h

Find the time “t” when their straight-line distance apart is the least.

QUESTION 2

A rectangle beam of width “w” and depth “d” is cut from a cylindrical pine log. The strength “s” of the beam is given by:

s = k(d^2)w where k is a constant.

Jonno is not sure whether to use a rectangle or square cross section

Determine which is stronger and justify reasoning

If i can get these two questions worked out it would be such a great help :D

Thanks

Answer 1

Scrander messed up.  He is usually pretty good.

x = 100 - 40t

y = 100 - 30 t

D = √[x^2 + y^2]

D = 10√[(100 - 80t + 16 t^2) + (100 - 60t + 9 t^2)

D = 10√[25t^2 -140t + 200]

If you square this, you can see write off that that t = 140/50

but you mise well do the derivative

d' = 0 = 10* [ (25t-70)/ √(25t^2 -140t + 200)]

t = 14/5

= 2 hr, 48 min

*********

You are finding when the minimum distance D, occurs.  This is when the derivative of the distance = 0

Answer 2

QUESTION 1

If you draw a triangle:

Let C = Caitlein's distance from intersection

Let J = John's distance from intersection

Distance between Caitlein and John = √(C + J)

J = 100 + 40t

C = 100 - 30t

D = √(100 - 30t + 100 + 40t)

D = √(200 + 10t)

d ' = 10 / 2√(200 + 10t)

Which never = 0 so the minimum must be at the start when t = 0

D =  Ã¢ÂˆÂš(200 + 10(0))

= 10√2km

QUESTION 2

s = k(d²)w

Strength will be maximised when s' = 0 but first we need to know how d and w are related

Because the log is cylindrical d and l are both points on a circle with radius r

let the middle of the circle be the point (0, 0)

the circle has equation:

w² + d² = r²

where w is the width and d is the height r is a constant

so d² = r² - w²

s = k(d²)w

s = k(r² - w²)w

s = kr²w - kw³

s ' = kr² - 3kw² = 0

kr² = 3kw²

r² = 3w²

w = r/√3

so when the width is r/√3

which means the depth is:

d² = r² - w²

d² = r² - (r/√3)²

d² = r² - r²/3

d² = 2r²/3

d = √(2)r/√3

so rectangular is stronger