HELP!! a particular river follows the quadratic curve f(x)=x^2, while a nearby road follow cubic...?

Question

A particular river follows the quadratic curve f(x)=x^2, while a nearby road follow cubic curve g(x)=ax^3+bx^2+cx+d.

There are three bridges on the road and they cross the river at the points (1,1), (0,0) and (1,1). The bridge at (1,1) is at right angles to the river. A dam is located at (1,2). The dam is not on this river. There is a straight spillway from the dam to the nearest point on the river.

a) What are the values of a,b,c,d?

b) What are the distances between teh bridges, as the crow flies?

c) At what angles (give both radians and degrees) do the other two bridges cross the river?

d) Where does teh spillway from the dam intersect the river?

e) A straight road is to be built from the existing road to the dam. Where on the original road should it start to have minimum length?



To anyone who could solve a part or all of this....thank you so much!!

supposedly there is enough detail. the question was given as a problem set for my ap calculus ab class.

Answer 1

f(x)=x^2, g(x)=ax^3+bx^2+cx+d.

i think the bridges should be at (-1,1), (0,0) and (1,1), since you typed (1,1) twice there.

qa
from (0,0)
0 = 0 + 0 + 0 + d, d = 0

from (-1,1) and (1,1)
1 = - a + b - c and
1 = a + b + c
2b = 2, b = 1
c = - a

f(x) = x^2
f'(x) = 2x
at (1,1), f'(1) = 2*1 = 2

and

g(x) = ax^3 +bx^2 +cx +d.
g'(x) = 3ax^2 + 2bx + c
= 3ax^2 + 2x - a

at (1,1), g'(1) should equal -1/2
-0.5 = 3a + 2 - a
2a = -5/2
a = -5/4
c = -a = 5/4

a = -5/4, b = 1, c = 5/4, d = 0

qb
if crow flies from 1 bridge to another, the pattern is a right-iso triangle.

distance of 2 legs
= (0,0) to (-1,1)
= (0,0) to (1,1)
= √2 unit

distance of hypo
= (-1,1) to (1,1)
= 2 unit (km or whatever)

qc
f'(x) = 2x
g'(x) = (-15x^2)/4 + 2x - 5/4

plug in x = -1 and x = 0
at (-1,1), f'(-1) = -2, g'(-1) = -7
at (0,0), f'(0) = 0, g'(0) = -5/4

angle at (-1,1)
= arctan (-2) - arctan (-7)
= ( find value )

angle at (0,0)
= arctan (0) - arctan (-5/4)
= -arctan (-5/4)
= ( find value )

qd
from f(x) = x^2,
intersection should be at P(p, p^2)

[d(p)]^2 = (p-1)^2 + (p^2 - 2)^2
= (p^2 -2p +1) + (p^4 - 4p^2 + 4)
= p^4 - 3p^2 - 2p + 5
d(p) = √(p^4 - 3p^2 - 2p + 5)

d'(p) = 0.5 * [(p^4 - 3p^2 - 2p + 5)^(-1/2)] * (4p^3 - 6p - 2)
= 0 when
2p^3 - 3p - 1 = 0
(p + 1)(2p^2 - 2p - 1) = 0
p = -1, p = (1 + √3)/2, p = (1 - √3)/2

p = (1 - √3)/2 lay in the middle, so it must be where the minimum value happen. or you could show with deriving d"(p) to show that.

so P((1 - √3)/2, (2 - √3)/2)

qe
Q(q, (-15q^2)/4 + 2q - 5/4)
dam(1,2)

[D(q)]^2 = (q-1)^2 + ((-15q^2)/4 + 2q - 5/4 - 2)^2
= (q^2 - 2q + 1) + ((-15q^2)/4 + 2q - 13/4)^2

differentiate D(q), find q when D(q) minimum (that's D'(q) = 0). just like the above.

Answer 2

d=0, since the road passes thru this point.
The bridges are sqrt(2) apart.
Is there enough data to do the problem?