Math 22 Exponential Growth and Decay - Revision history

Tphan046 at 15:18, 12 August 2020

2020-08-12T15:18:05Z

2020-08-12T14:59:15Z

2020-08-12T14:58:46Z

2020-08-12T14:48:41Z

2020-08-11T17:16:49Z

2020-08-11T17:16:06Z

Guidelines for Modeling Exponential Growth and Decay

2020-08-11T17:15:59Z

2020-08-11T17:11:51Z

Exponential Growth and Decay

2020-08-11T17:11:39Z

2020-08-11T17:09:59Z

Created page with " '''Return to Topics Page''' '''This page were made by Tri Phan'''"

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[[Math_22| '''Return to Topics Page''']]

'''This page were made by [[Contributors|Tri Phan]]'''

← Older revision		Revision as of 15:18, 12 August 2020
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−	~~Under construction~~
	==Exponential Growth and Decay==		==Exponential Growth and Decay==
	If <math>y</math> is a positive quantity whose rate of change with respect to time		If <math>y</math> is a positive quantity whose rate of change with respect to time

@@ Line 31: / Line 31: @@
 |}
-'''b)''' A sports utility vehicle that costs $33000 new has a book value of $20856 after 3 years. Assume the value of the vehicle exponential decay over time. Find this exponential model.
+'''2)''' A sports utility vehicle that costs $33000 new has a book value of $20856 after 3 years. Assume the value of the vehicle exponential decay over time. Find this exponential model.
 {| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 !Solution: &nbsp;

@@ Line 31: / Line 31: @@
 |}
-'''b)''' <math>f(x)=\ln (x^8)</math>
+'''b)''' A sports utility vehicle that costs $33000 new has a book value of $20856 after 3 years. Assume the value of the vehicle exponential decay over time. Find this exponential model.
 {| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 !Solution: &nbsp;
 |-
-|Solution 1: <math>f'(x)=\frac{1}{x^8}(x^8)'=\frac{8x^7}{x^8}=\frac{8}{x}</math>
+|Consider: <math>y=Ce^{kt}</math>
 |-
-|Solution 2: <math>f(x)=\ln (x^8)=8\ln x</math>, so <math>f'(x)=8\frac{1}{x}=\frac{8}{x}</math>
+|So, <math>20856=33000e^{k3}</math>, then <math>\frac{20856}{33000}=\frac{79}{125}=e^{3k}=(e^k)^3</math>
 |}

← Older revision		Revision as of 14:48, 12 August 2020
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	3. Use the model <math>y=Ce^{kt}</math> to answer the question.		3. Use the model <math>y=Ce^{kt}</math> to answer the question.

		+	'''Exercises'''
		+
		+	'''1)''' The number of a certain type of bacteria increases continuously at a rate proportional to the number present. There are 150 bacteria at a given time and 450 bacteria 5 hours later. How many bacteria will there be 10 hours after the initial time?
		+	{\| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
		+	!Solution:
		+	\|-
		+	\|At the initial time: <math>y_0=Ce^{kt_0}</math>, so <math>150=Ce^{kt_0}</math>
		+	\|-
		+	\|After 5 hours: <math>y_1=Ce^{k(t_0+5)}</math>, so <math>450=Ce^{k(t_0+5)}=Ce^{kt_0+5k}=Ce^{kt_0}e^{5k}</math>
		+	\|-
		+	\|Take the second equation and divided it by the first equation, we get:
		+	\|-
		+	\|<math>\frac{450}{150}=\frac{Ce^{kt_0}e^{5k}}{Ce^{k(t_0+5)}}</math>
		+	\|-
		+	\|<math>3=e^{5k}</math>
		+	\|-
		+	\|Now, consider the equation after 10 hours: <math>y_2=Ce^{k(t_0+10)}=Ce^{k[(t_0+5)+5]}=Ce^{k(t_0+5)}e^{kt}=450.3=1350</math>
		+	\|}
		+
		+	'''b)''' <math>f(x)=\ln (x^8)</math>
		+	{\| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
		+	!Solution:
		+	\|-
		+	\|Solution 1: <math>f'(x)=\frac{1}{x^8}(x^8)'=\frac{8x^7}{x^8}=\frac{8}{x}</math>
		+	\|-
		+	\|Solution 2: <math>f(x)=\ln (x^8)=8\ln x</math>, so <math>f'(x)=8\frac{1}{x}=\frac{8}{x}</math>
		+	\|}

← Older revision		Revision as of 17:16, 11 August 2020
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		+	Under construction
	==Exponential Growth and Decay==		==Exponential Growth and Decay==
	If <math>y</math> is a positive quantity whose rate of change with respect to time		If <math>y</math> is a positive quantity whose rate of change with respect to time

@@ Line 7: / Line 7: @@
 ==Guidelines for Modeling Exponential Growth and Decay==
 . Use the given information to write two sets of conditions involving <math>y</math> and <math>t</math>
-. Substitute the given conditions into the model <math>y=Ce^{kt}</math> and use the results to solve for the constants <math>C</math> and <math>k</math>.
+. Substitute the given conditions into the model <math>y=Ce^{kt}</math> and
 . Use the model <math>y=Ce^{kt}</math> to answer the question.

@@ Line 5: / Line 5: @@
    where <math>C</math> is the initial value and <math>k</math> is the constant of proportionality.
    Exponential growth is indicated by <math>k>0</math> and exponential decay by <math>k<0</math>.
 [[Math_22| '''Return to Topics Page''']]
 '''This page were made by [[Contributors|Tri Phan]]'''

← Older revision		Revision as of 17:11, 11 August 2020
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		+	==Exponential Growth and Decay==
		+	If <math>y</math> is a positive quantity whose rate of change with respect to time
		+	is proportional to the quantity present at any time <math>t</math>, then is of the form
		+	<math>y=Ce^{kt}</math>
		+	where <math>C</math> is the initial value and <math>k</math> is the constant of proportionality. Exponential growth is indicated by <math>k>0</math> and exponential decay by <math>k<0</math>.
	[[Math_22\| '''Return to Topics Page''']]		[[Math_22\| '''Return to Topics Page''']]

	'''This page were made by [[Contributors\|Tri Phan]]'''		'''This page were made by [[Contributors\|Tri Phan]]'''