palled into a dock by a rope attached to the bow of the boat and passing through

Question

palled into a dock by a rope attached to the bow of the boat and passing through a pulley Ar deck that is Inn higher than the bow of the boat. if the rope pulled in at the rate of 1m/s fat the boat is approaching the dock when it is 8m from the dock? 6. If a snowball melts so that its surface area decreases at the rate of 1cm /min, find the rate at which the diameter decreases when the diameter is 10am. (Area of a Sphere A 7, The altitude of a triangle is increasing at a rate of icm/min while the area of the triangle is increasing at a rate of /min. At what rate is the base of the triangle is changing when the altitude is 10cm and the area is 100cm

Explanation / Answer

6: Solution:

We know that the surface area of a (spherical) snowball is given by
A = 4r2 = d2    ( r is radius and d is diameter )

and that the diameter is 2r

Also, we know that

dA/dt = -1 ( Given)

and that, at the time of interest, the diameter is 10 cm, meaning that r = 5.

Our goal is to determine the rate at which the diameter decreases,

which will be 2(dr/dt)

differentiating dA/dt = 4 (2r)dr/dt = 8r dr/dt = -1

Now plug r = 5

we get

8*5* dr/dt = -1

2dr/dt = -1/20

2dr/dt = rate chage of diameter = -1/20

the diameter is decreasing at a rate of 1/20 = 0.016 cm / min

Answer

7.

The area of a triangle is A(t) = 1/2 H(t) * B(t),

where H(t) denotes the hight of the triangle at
time t and B(t) denotes the size of the base.

Then the area varies as follows

:A'(t) = 1/2 (H'B +HB')

B' = ((2A' (t) -H' B) / H)

We have

A' = 2cm/min and H' = 1cm/min;

thus plug this value into above

we get

B' = (2.2 - 1.2 ) / 10 = -1.6 cm / min,

i.e. the base is decreasing by 1.6 cm/min

Answer

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