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The volume and surface area of a sphere both depend on its radius. = V - πr³, S = 4πr². (i) Find the rate of change of the volume with respect to the radius and the rate of change of the surface area with respect to the radius. dV dr = 4πr² dS = 8πr dr (ii) Find the rate of change of the surface area to volume ratio S/V with respect to the radius. d(S/V) dr = -3/r² (iii) Eliminate the radius and express V as a function of S. V == (1/6г^(1/2))S^(3/2) (iv) Find the rate of change of the volume with respect to the surface area. dV = r/2 cm dS cm In this question, we will evaluate the limit 1077 x 1 lim x-1 x (a) Is the function x1077 1 x- 1 continuous at x = 1? What does that mean for computing the limit? The function is not continuous at x=1, so the limit automatically does not exist. (b) Make the substitution h = = x - 1. What happens to the limit? 1077 х - lim x+1 x- 1 1 = lim lim(h→0) ((h+1)^1077 - 1)/h h→0 (c) Your limit in (b) looks a lot like a derivative. Find a function f(x) and a constant a so that your limit above computes f'(a). Then, using derivative rules, evaluate the limit. lim x→1 1077 x - x-1 - 1 = = f'(a) = 1

The volume and surface area of a sphere both depend on its radius. = V - πr³, S = 4πr². (i) Find the rate of change of the volume with respect to the radius and the rate of change of the surface area with respect to the radius. dV dr = 4πr² dS = 8πr dr (ii) Find the rate of change of the surface area to volume ratio S/V with respect to the radius. d(S/V) dr = -3/r² (iii) Eliminate the radius and express V as a function of S. V == (1/6г^(1/2))S^(3/2) (iv) Find the rate of change of the volume with respect to the surface area. dV = r/2 cm dS cm In this question, we will evaluate the limit 1077 x 1 lim x-1 x (a) Is the function x1077 1 x- 1 continuous at x = 1? What does that mean for computing the limit? The function is not continuous at x=1, so the limit automatically does not exist. (b) Make the substitution h = = x - 1. What happens to the limit? 1077 х - lim x+1 x- 1 1 = lim lim(h→0) ((h+1)^1077 - 1)/h h→0 (c) Your limit in (b) looks a lot like a derivative. Find a function f(x) and a constant a so that your limit above computes f'(a). Then, using derivative rules, evaluate the limit. lim x→1 1077 x - x-1 - 1 = = f'(a) = 1

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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The volume and surface area of a sphere both depend on its radius. = V - πr³, S = 4πr². (i) Find the rate of change of the volume with respect to the radius and the rate of change of the surface area with respect to the radius. dV dr = 4πr² dS = 8πr dr (ii) Find the rate of change of the surface area to volume ratio S/V with respect to the radius. d(S/V) dr = -3/r² (iii) Eliminate the radius and express V as a function of S. V == (1/6г^(1/2))S^(3/2) (iv) Find the rate of change of the volume with respect to the surface area. dV = r/2 cm dS cm
Transcribed Image Text:The volume and surface area of a sphere both depend on its radius. = V - πr³, S = 4πr². (i) Find the rate of change of the volume with respect to the radius and the rate of change of the surface area with respect to the radius. dV dr = 4πr² dS = 8πr dr (ii) Find the rate of change of the surface area to volume ratio S/V with respect to the radius. d(S/V) dr = -3/r² (iii) Eliminate the radius and express V as a function of S. V == (1/6г^(1/2))S^(3/2) (iv) Find the rate of change of the volume with respect to the surface area. dV = r/2 cm dS cm
In this question, we will evaluate the limit 1077 x 1 lim x-1 x (a) Is the function x1077 1 x- 1 continuous at x = 1? What does that mean for computing the limit? The function is not continuous at x=1, so the limit automatically does not exist. (b) Make the substitution h = = x - 1. What happens to the limit? 1077 х - lim x+1 x- 1 1 = lim lim(h→0) ((h+1)^1077 - 1)/h h→0 (c) Your limit in (b) looks a lot like a derivative. Find a function f(x) and a constant a so that your limit above computes f'(a). Then, using derivative rules, evaluate the limit. lim x→1 1077 x - x-1 - 1 = = f'(a) = 1
Transcribed Image Text:In this question, we will evaluate the limit 1077 x 1 lim x-1 x (a) Is the function x1077 1 x- 1 continuous at x = 1? What does that mean for computing the limit? The function is not continuous at x=1, so the limit automatically does not exist. (b) Make the substitution h = = x - 1. What happens to the limit? 1077 х - lim x+1 x- 1 1 = lim lim(h→0) ((h+1)^1077 - 1)/h h→0 (c) Your limit in (b) looks a lot like a derivative. Find a function f(x) and a constant a so that your limit above computes f'(a). Then, using derivative rules, evaluate the limit. lim x→1 1077 x - x-1 - 1 = = f'(a) = 1
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