x^2 +y^2 =s^2
2xdx/dt +2ydy/dt = 2sds/dt
dy/dt = 60
lynn has gone 10 miles so t = 1/5
now calculate charles' distance using 1/5
calculate s from original equation
plug and chug
I am frustrated to death with this problem and would really appreciate some guidane.
Lynn and Charlie work in offices that are 40 miles apart. Lynn leaves home and drives to her office going due east at a rate of 50 mph. At the same time, charleie leaves home and drives to his office going due north at a rate of 60mph. How fast is the distance between Lynn and Charlie changing when Lynn has driven 10 miles.
The diagram in the problem shows that it's supposed to use pythagoran theorem. The people both leave fromo practically the same spot so it makes a right angle triangle.
Thanks very much
at 1/5 of hr they are 15.6 miles apart. If they are leaving from home the distances their offices are apart seems irrelevant other than that neither one had already reached their office after 1/5 hr--So I have to agree with awkward have you given us the whole problem?