Let A be the area of a circle with radius r. If dr/dt = 3, find dA/dt when r = 1

Question

Let A be the area of a circle with radius r. If dr/dt = 3, find dA/dt when r = 1. A spherical snowball is melting in such a way that its diameter is decreasing at rate of 0.2 cm/min. At what rate is the volume of the snowball decreasing when the diameter is 12 cm. (Note the answer is a positive number). cm^3/min A street light is at the top of a 16 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 8 ft/sec along a straight path. How fast is the tip of her shadow moving when she is 35 ft from the base of the pole? ft/sec Note: You should draw a picture of a right triangle with the vertical side representing the pole, and the other end of the hypotenuse representing the tip of the woman's shadow. Where does the woman fit into this picture? Label her position as a variable, and label the tip of her shadow as another variable. You might like to use similar triangles to find a relationship between these two variables. Water is leaking out of an inverted conical tank at a rate of 14300.0 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 9.0 meters and the diameter at the top is 4.0 meters. If the water level is rising at a rate of 27.0 centimeters per minute when the height of the water is 1.5 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute. Note: Let "R" be the unknown rate at which water is being pumped in. Then you know that if V is volume of water, dV/dt = R - 14300.0. Use geometry (similar triangles?) to find the relationship between the height of the water and the volume of the water at any given time. Recall that the volume of a cone with base radius r and height h is given by 1/3 pi r^2 h.

Explanation / Answer

a)area A =r2

differentiate with respect to time t

dA/dt = 2r(dr/dt)

given dr/dt =3, r=1

dA/dt = 2*1*(3)

dA/dt = 6

dA/dt = 18.85

2)volume of spherical ball V=(4/3)r3, where r is radius, radius =diameter/2 =d/2

V=(4/3)(d/2)3

V=(1/6)d3

differentiate with respect to time t

dV/dt =(1/6)3d2(dd/dt)

dV/dt =(1/2)d2(dd/dt)

given diameter decreasing at rate of 0.2cm/min =>dd/dt =-0.2, diameter =12 cm

dV/dt =(1/2)122(-0.2)

dV/dt =-14.4

dV/dt =-45.24

volume decreasing at rate of 45.24 cm3/min

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