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
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
2 answers
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
1 answer