You need to pay attention to the sign +/- of the derivatives you're using. Velocities are vectors and vectors have directions. Did you draw a diagram and set up a coordinate system?
Stuck on this one.
An airplane leaves a field at noon and flies east at 100 km/h and a second airplane leaves the same field at 1pm and flies south at 150 km/h. How fast are the airplanes separating at 2pm.
So if I do a right triangle.
My Vx right would be 200 and x would be 2 for hours//
My Vy(south) would be 150 and y 1 hour.
My Resultant of Vz is what I am looking for and z is 2.23606.
I am to find Vz and I come up with 245 km/h and the answer is 170 km/h.
Can someone see what I did wrong?? Thanks
Joanne
Oh, ok. Maybe you can avoid using vectors, and just look at it as a triangle. To find the position of the plane directed east, you need , where t is in hours. The distance of the second plane will be one hour behind, so it's position will be . The distance function could be written:
You're given the velocities:
So you plug those into the formula, and expand to obtain:
Use the chainrule to differentiate this function in order to find the rate at which the distance between the planes is increasing:
This derivative is equal to 170 at t=2.