AP Calculus ABhardmcq1 pt

At noon, ship A is 100 km west of ship B. Ship A sails east at 20 km/h and ship B sails north at 15 km/h. How fast is the distance between them changing at 2 PM?

A.2525 km/h, decreasing
B.1515 km/h, decreasing
C.3535 km/h, increasing
D.555\sqrt{5} km/h, decreasing

Explanation

Core Concept

Let zz be the distance. From z2=(10020t)2+(15t)2z^2=(100-20t)^2+(15t)^2, at t=2t=2 hours, z=305z=30\sqrt{5} and zdzdt=60(20)+30(15)=750z\dfrac{dz}{dt}=60(-20)+30(15)=-750. Thus dzdt=750305=55\dfrac{dz}{dt}=-\dfrac{750}{30\sqrt{5}}=-5\sqrt{5}, so the distance decreases at 555\sqrt{5} km/h.

Correct Answer

D555\sqrt{5} km/h, decreasing

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