Field conditions require a highway curve to pass through a fixed point. Compute a suitable equal-tangent...

Field conditions require a highway curve to pass through a fixed point. Compute a suitable equal-tangent vertical curve and full-station elevations for grades of g1 = −2.50% and g2 = +1.00%, VPI elevation 750.00 ft at station 30 + 00. Fixed elevation 753.00 ft at station 30 + 00.

Expert Solution

Ally Wells answered 9 months ago

Find the equation of the tangent line to the curve at the point corresponding to the...

Find the equation of the tangent line to the curve at the point corresponding to the given value of t 1. x=cost+tsint, y=sint-tcost t=7pi/4 2. x=cost+tsint, y=sint-tcost t=3pi/4?

(a) Suppose that the tangent line to the curve y = f (x) at the point...

(a) Suppose that the tangent line to the curve y = f (x) at the point (−9, 53) has equation y = −1 − 6x. If Newton's method is used to locate a root of the equation f (x) = 0 and the initial approximation is x1 = −9, find the second approximation x2. (b) Suppose that Newton's method is used to locate a root of the equation f (x) = 0 with initial approximation x1 = 9. If the...

For the 405 highway that car pass through a checkpoint, assume the speeds are normally distributed...

For the 405 highway that car pass through a checkpoint, assume the speeds are normally distributed such that μ= 61 miles per hour and δ=4 miles per hour. Calculate the Z value for the next car that passes through the checkpoint will be traveling slower than 65 miles per hour. Calculate the Z value for the next car that passes through the checkpoint will be traveling more than 66 miles per hour. Calculate the probability that the next car will...

How do you find the slope of the tangent line to a curve at a point?

What is the equation of the line tangent to the curve at point (12,5)? Sorry! Forgot...

What is the equation of the line tangent to the curve at point (12,5)? Sorry! Forgot to include the curve equation y = 3x2 - 2x + 1

Find the equation of the tangent line to the curve at the given point using implicit...

Find the equation of the tangent line to the curve at the given point using implicit differentiation. Cardioid (x2 + y2 + y)2 = x2 + y2 at (−1, 0)

Find the slope of the tangent line to the following curve at the point given. r=...

Find the slope of the tangent line to the following curve at the point given. r= Cos(3theta) + Sin(2theta) (1,0)

Use the following data to calculate the layout of an equal tangent vertical curve by offsets...

Use the following data to calculate the layout of an equal tangent vertical curve by offsets for every full station. g1 = -1.20% g2 = +3.50% X PVI = 49+70 Y PVI = 5,893.48’ L = 1500’ , find the station and elevation of the high (or low) point of the curve.

1.)The perpendicular distance from the tangent at PC to a point Q on a simple curve...

1.)The perpendicular distance from the tangent at PC to a point Q on a simple curve is 64m, If the distance from PC to Q is 260m, determine the station of Q when PC is at STA 12+598. 2.) Two parallel roads 25m apart are to be connected by a reversed curve with equal radii at PC and PT. The length of the common radius is 1250m.apart are to be connected by a reversed curve with equal radii at PC...

An equal-tangent crest vertical curve connects an initial grade of +2.5% and a final grade of...

An equal-tangent crest vertical curve connects an initial grade of +2.5% and a final grade of –0.5%. The curve is designed for 70 mi/h and the station of the PVT is 132+62 and is at elevation 833 ft. What is the station and elevation of the curve's high point?

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