PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)
PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)
8th Edition
ISBN: 9781337904285
Author: Larson
Publisher: CENGAGE L
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Chapter 9.1, Problem 98E

(a)

To determine

To find:The equation of the parabola with its vertex at origin that models the road surface.

(a)

Expert Solution
Check Mark

Answer to Problem 98E

The equation of the parabola is x2=640y .

Explanation of Solution

Given information:

Roads are often designed with parabolic surfaces to allow rain to drain off. A particular road that is 32 feet wide is 0.4 feet higher in the center than it is on the sides.

  PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS), Chapter 9.1, Problem 98E , additional homework tip  1

Calculation:

As clear from the figure given above the parabola is vertically downwards with vertex at origin and also it passes through the point (16,0.4) .

The general equation of the parabola is x2=4py .

Now substitute 16 for x and 0.4 for y in the equation.

  x2=4py162=4p(0.4)p=2560.16p=160

Now substitute 160 for p in the equation of the parabola.

  x2=4pyx2=4(160)yx2=640y

Therefore, the equation of the parabola is x2=640y .

(b)

To determine

To calculate:The distance of the surface from the center of the road at the time of surface being 0.1 foot lower than in the middle.

(b)

Expert Solution
Check Mark

Answer to Problem 98E

The distance of the surface from the center of the road at the time of surface being 0.1 foot lower than in the middle is 13.86 foot.

Explanation of Solution

Given information:

Roads are often designed with parabolic surfaces to allow rain to drain off. A particular road that is 32 feet wide is 0.4 feet higher in the center than it is on the sides.

  PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS), Chapter 9.1, Problem 98E , additional homework tip  2

Calculation:

As calculated in part(a), the equation of the parabola is x2=640y .

Now the height of the surface is 0.1 foot lower than in the middle.

So,

  y=0.40.1=0.3 .

Now substitute 0.3 for y in the equation of the parabola.

  x2=640yx2=640(0.3)x2=192x=±13.86

Therefore, the distance of the surface from the center of the road at the time of surface being 0.1 foot lower than in the middle is 13.86 foot.

Chapter 9 Solutions

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

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Prob. 24CTCh. 9 - Prob. 25CTCh. 9 - Prob. 1STPCh. 9 - Prob. 2STPCh. 9 - Prob. 3STPCh. 9 - Prob. 4STPCh. 9 - Prob. 5STPCh. 9 - Prob. 6STPCh. 9 - Prob. 7STPCh. 9 - Prob. 8STPCh. 9 - Prob. 9STPCh. 9 - Prob. 10STPCh. 9 - Prob. 11STPCh. 9 - Prob. 12STPCh. 9 - Prob. 13STPCh. 9 - Prob. 14STPCh. 9 - Prob. 15STPCh. 9 - Prob. 16STPCh. 9 - Prob. 17STPCh. 9 - Prob. 18STPCh. 9 - Prob. 19STPCh. 9 - Prob. 20STPCh. 9 - Prob. 21STPCh. 9 - Prob. 22STPCh. 9 - Prob. 23STPCh. 9 - Prob. 24STPCh. 9 - Prob. 25STPCh. 9 - Prob. 26STPCh. 9 - Prob. 27STP
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