A rectangular barge is traveling under a bridge with a parabolic archway. The barge is 60 feet tall and 80 feet wide. The bridge is 80 feet tall and 200 feet wide.
If the barge must travel down the right side of the river to allow two-way traffic, will it fit under the bridge? How do you know?
What if the barge is allowed to travel down the center of the river?
I've never done a word problem containing parabolas before, so I'm not sure how to approach this.
Choose your coordinates first. Here it seems natural for $x$ to be horizontal with $0$ the center of the river and $y$ to be vertical with $0$ the water surface. Because we centered it, the equation of the bridge will be $y=a-bx^2$ use the points you are given to find $a,b$. Then see if you put one side of the ship on the centerline of the river if it fits and then if you put the center of the ship on centerline it fits. I gave my guess in the comments.
Solution : It must be clear from then diagram. The equation of the parabola = $y = -bx^2$
Find b,
Now using b evaluate Y1 at x = 80 and you will see te clearance available is only 28.8 which is less than the 60 m required for the barge to travel to the right and allow two way traffic
Now using b evaluate Y2 at x = 40 and you will see the clearance available is 67.2 m which is more than the 60 m required for the barge to travel in the center.
Thanks
Satish
General equation of parabola is $y=a(x-h)^2 +k$. For the bridge, the vertex K is 80ft and the axis of symmetry is set at $x=0$, which means $h = 0$.
Thus $y=ax^2+80$. To solve for $a$, plug in $x$ and $y$. When $y=0$, $x= \pm 100$. Thus $0=a(100)^2 + 80$ and $a=-\frac{2}{125} ft$.
Thus $y=-\frac{2}{125} x^2 + 80$.
The barge is 60 ft tall and 80 ft wide (presumably these are heights above the water line).
In the parabolic equation what is span/clearance x when y=60 ft?
$60=-\frac{2}{125} x^2 + 80$....solving for x, we get $x= \pm 50$. That means at a height of 60 ft, the bridge has a width of 100 feet. This gives a clearance of 20 ft for the barge.
DK
