Wednesday

November 26, 2014

November 26, 2014

Posted by **Kristy** on Wednesday, November 30, 2011 at 8:15pm.

- Calculus -
**bobpursley**, Wednesday, November 30, 2011 at 8:27pmdraw the right triangle.

start the origin at B at noon.

a is the distance ship A moves in 4 hours.

a= -180+40*4=-20km

da/dt= + 40

b is the distance from the B starting point at noon.

b= 30*4=120

db/dt=30\

r=sqrt(a^2+b^2)

dr/dt= 1/2 *1/(a^2+b^2)* (2a da/dt+2b db/dt)

solve for dr/dt

- Calculus -
**Kristy**, Thursday, December 1, 2011 at 10:14pmI tried plugging in the numbers for the variables, but I am not coming up with the right answer. I think the equation might be wrong.

**Answer this Question**

**Related Questions**

calculus - At noon, ship A is 150 km west of ship B. Ship A is sailing east at ...

calculus - At noon, ship A is 150 km west of ship B. Ship A is sailing east at ...

Calculus - At noon, ship A is 150 km west of ship B. Ship A is sailing east at ...

calculus - At noon, ship A is 110 km west of ship B. Ship A is sailing east at ...

Calculus - "At 9am ship A is 50 km east of ship B. Ship A is sailing north at ...

calculus - At noon, ship A is 100 kilometers due east of ship B. Ship A is ...

Calc - At noon, ship A is 100km west of ship B. Ship A is sailing south at 30km/...

Calculus 1 - Related Rates I am having trouble finding the equation to use to ...

Calculus 1 - Related Rates I am having trouble finding the equation to use to ...

Calculus 1 - Related Rates I am having trouble finding the equation to use to ...