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Suppose that the volume V of a rolling snowball increases so that is proportional to the surface area of the snowball at time t. Show that the radius r increases at a constant rate, that is, dr = 42 dr 2 d dt The volume of the snowball is given by V=²³, so d Since the change in volume is proportional to the surface area S, with S= 4m², we also have the following for some constant k. dv O = r. 4mk² O O 92 92 92 = k-4tr² = k·m² •r. Tk³ = k· 2mr Equating the two expressions for gives 4m² 2 dr - k dt => dt =k, that is, dt is constant. is constant.

Suppose that the volume V of a rolling snowball increases so that is proportional to the surface area of the snowball at time t. Show that the radius r increases at a constant rate, that is, dr = 42 dr 2 d dt The volume of the snowball is given by V=²³, so d Since the change in volume is proportional to the surface area S, with S= 4m², we also have the following for some constant k. dv O = r. 4mk² O O 92 92 92 = k-4tr² = k·m² •r. Tk³ = k· 2mr Equating the two expressions for gives 4m² 2 dr - k dt => dt =k, that is, dt is constant. is constant.

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Suppose that the volume V of a rolling snowball increases so that The volume of the snowball is given by V = = π²³, so dy = - T O O Since the change in volume is proportional to the surface area S, with S = 4m², we also have the following for some constant k. dv O dt dv dk = r. 4nk² = k· 4tr² = k· πr³ = r. πk³ = k 2mr Equating the two expressions for dv dt is proportional to the surface area of the snowball at time t. Show that the radius r increases at a constant rate, that is, dt 2 dr = k gives 42 dr dt = 4tr² dr dt =k, that is, dt dt is constant. is constant.
Transcribed Image Text:Suppose that the volume V of a rolling snowball increases so that The volume of the snowball is given by V = = π²³, so dy = - T O O Since the change in volume is proportional to the surface area S, with S = 4m², we also have the following for some constant k. dv O dt dv dk = r. 4nk² = k· 4tr² = k· πr³ = r. πk³ = k 2mr Equating the two expressions for dv dt is proportional to the surface area of the snowball at time t. Show that the radius r increases at a constant rate, that is, dt 2 dr = k gives 42 dr dt = 4tr² dr dt =k, that is, dt dt is constant. is constant.
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