Answered

Engage with knowledgeable experts and get accurate answers on IDNLearn.com. Ask your questions and get detailed, reliable answers from our community of experienced experts.

One shape is approaching a poor from the east traveling west at 15 mph and its presently 3 miles east east of the port I second ship is already left the poor traveling to the north at 10 mph and is presently 4 miles north of the port at this instant what is the rate of change of distance between the two ships are they going to close your more for their apart

Sagot :

Answer:

DL/dt = 17 mph

Step-by-step explanation:

The ships and the port shape a right triangle

Ship  going west ( x-direction)   is traveling at 15 mph

Ship going north  (y-direction) is traveling at 10 mph

The distance L between ships is:

L²  = x² + y²

Tacking derivatives on both sides of the equation with respect to time

we get

2*L*DL/dt = 2*x*Dx/dt  + 2*y*Dy/dt    (1)

In that equation we know:

Dx/dt  = 15 mph

Dy/dt  = 10 mph

At the moment  ship traveling from the east is at 3 miles from the port

then x = 3 m  and the other ship is at 4 miles north

then by Pythagoras theorem

L = √ 3² + 4²    L = 5

By substitution in equation 1

2*5*DL/dt  = 2*3*15 + 2*4*10

10* DL/dt  =  90  +  80

DL/dt = 170 / 10

DL/dt = 17 mph

We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. IDNLearn.com is your go-to source for dependable answers. Thank you for visiting, and we hope to assist you again.