A) The altitude (i.e., height) of a triangle is increasing at a rate of 2 cm/min

Question

A) The altitude (i.e., height) of a triangle is increasing at a rate of 2 cm/minute while the area of the triangle is increasing at a rate of 4.5 square cm/minute. At what rate is the base of the triangle changing when the altitude is 7.5 centimeters and the area is 86 square centimeters? The base is changing at _________ cm/min.

B) A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1 m higher than the bow of the boat.

If the rope is pulled in at a rate of 1.2 m/s, how fast is the boat approaching the dock when it is 10 m from the dock? Rate=_______ m/s

C) At noon, ship A is 20 nautical miles due west of ship B. Ship A is sailing west at 22 knots and ship B is sailing north at 16 knots. How fast (in knots) is the distance between the ships changing at 4 PM? The distance is changing at  knots. (Note: 1 knot is a speed of 1 nautical mile per hour.)

Explanation / Answer

a)let base of triangle =b ,height =h , area =A

area of triangle A=(1/2)* b*h

Differentiate with respect to trime on both sides .product rule;(uv)'=u'v +uv'

dA/dt =(1/2)(db/dt)h +(1/2)bdh/dt

given height h=7.5cm, area A=86cm2

A=(1/2)* b*h

86= (1/2)* b*7.5

b =344/15

base =344/15cm

height of a triangle is increasing at a rate of 2 cm/min=> dh/dt =2

the area of the triangle is increasing at a rate of 4.5 square cm/min=> dA/dt =4.5

dA/dt =(1/2)(db/dt)h +(1/2)bdh/dt

4.5 =(1/2)(db/dt)7.5 +(1/2)(344/15)*2

4.5 =(15/4)(db/dt) +(344/15)

(15/4)(db/dt) =4.5-(344/15)

(db/dt) =(4/15)[4.5-(344/15)]

db/dt=-4.915

The base is changing at 4.195 cm/min.

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