Asked by diego
a road that runs perpendicular to a highway leads to a farmhouse located one mile off the highway an automobile travels down the highway past the road leading to the farmhouse at a speed of 60 mph how fast is the distance between the farmhouse an automobile increasing when the car is 3 miles past the intersection of the highway and the road to the farmhouse
Answers
Answered by
Reiny
It would help if you posted your question using proper English, with capitals and periods.....
At a time of t hours, let the distance of the car past the intersection be x miles.
let the distance between the car and the farmhouse be d
I get a simple right-angled triangle with
sides 1, x and hypotenuse d
d^2 = 1^2 + x^2
2d dd/dt = 2x dx/dt
or dd/dt = x dx/dt /d
when x = 3
d^2 = 1 + 9 = 10
d =√10 , and we are also given that dx/dt = 60
dd/dt = 3(60)/√10 = 180/√10 = appr 56.9 mph
At a time of t hours, let the distance of the car past the intersection be x miles.
let the distance between the car and the farmhouse be d
I get a simple right-angled triangle with
sides 1, x and hypotenuse d
d^2 = 1^2 + x^2
2d dd/dt = 2x dx/dt
or dd/dt = x dx/dt /d
when x = 3
d^2 = 1 + 9 = 10
d =√10 , and we are also given that dx/dt = 60
dd/dt = 3(60)/√10 = 180/√10 = appr 56.9 mph
Answered by
Steve
if the car is x miles past the road, the distance is
s^2 = 1+x^2
when x=3, s=√10
2s ds/dt = 2x dx/dt
ds/dt = x/s dx/dt
when x=3, then,
ds/dt = 3/√10 * 60 = 18√10 = 56.9 mph
s^2 = 1+x^2
when x=3, s=√10
2s ds/dt = 2x dx/dt
ds/dt = x/s dx/dt
when x=3, then,
ds/dt = 3/√10 * 60 = 18√10 = 56.9 mph
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