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8 Cards in this Set
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A rock dropped into a pond causes a circular wave of ripples whose radius increases at 4 inches per second. How fast is the area of the circle of ripples expanding at the instant that the radius of the circle is 12 inches? 24 inches? 100 inches? Explain why it makes sense that the rate of change of the area increases as the radius increases |
A = pi*r^2 dA/dt = 2pir*dr/dt dA/dt = 2*pi*12*4 = 96pi |
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1)Supposes the sides of a cube are expanding at a rate of 2 inches per minute How fast is the volume of the cube changing at the moment that the cube has a side length of 8 inches? |
x = 8 V = x^3 dV/dt = 3x^2*dx/dt dV/dt = 3(8)^2*2 = 384 cubic inches / s |
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2)How fast is the volume of the cube changing at the moment that the cube's volume is 55 cubic inches? |
V = x^3 55 = x^3 x = 3sqrt(55) dV/dt = 3x^2*dx/dt = 3(3sqrt(55))^2*2 = 86.7 cubic inches |
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Consider a large helium baloon that is being inflated at the rate of 120 cubic inches per second: 1) How fast is the radius of the balloon increasing at the instant that the balloon has a radius of 12 inches? |
r = 12 dV/dt = 120 V = 4/3*pi*r^3 dV/dt = 4*pi*r^2*dr/dt 120 = 576pi*dr/dt dr/dt = 120/576pi inches / s |
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2) How fast is the surface area of the balloon increasing the instant that the radius of the balloon is 15 inches? |
r = 15 dV/dt = 120 V = 4/3*pi*r^2 dV/dt = 4*pi*15^2*dr/dt dr/dt = 120/900pi S = 4*pi*r^2 dS/dt = 8*pi*r* (120/900pi) dS/dt = 120*pi*(120/900pi) dS/dt = 16 |
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Linda is bored and decides to pour an entire container of salt into a pile on the kitchen floor. She pours 3 cubic inches of salt per second into a conical pile whose height is always two third its radius. 1) How fast is the radius of the conical salt pile changing when the height is 4 inches |
h=2/3r r = 3/2*h dv/dt = 3 r = 3/2*4 = 6 V = 1/3*pi*r^3*2/3 V = 2/9*pi*r^3 dV/dt = 2/3*pi*r^2*dr/dt 3 = 24pi*dr/dt dr/dt = 3/24pi inches / s |
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2) How fast is the height of the conical salt pile changing when the height is 4 inches |
r = 3/2*h V = 1/3*pi*(3/2*h)^2*h v = 3/4*pi*h^3 dv/dt = 9/4*pi*h^2*dh/dt 3 = 9/4*pi*4^2*dh/dt dh/dt = 3/36pi = 1/12pi inches / s |
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Riley is holding an ice cream cone on a hot summer day. The cone has a small hole at the bottom and ice cream is melting and dripping through the hole at a rate of half a cubic inche per second. The cone has a radius of 2 inches and a height of 5 inches. How fast is the height of the ice cream changing when the height of the ice cream in the cone is 3 inches |
r = 2/5h h = 3 dv/dt = -0.5 V = 1/3*pi*r^2*h V = 1/3*pi*(2/5h)^2*h V = 4/75*pi*h^3 dV/dt = 4/25*pi*h^2*dh/dt -0.5 = 4/25*pi*3^2*dh/dt dh/dt = -25/72pi inches / s |
