Difference between revisions of "Math 22 Natural Exponential Functions"

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   Let <math>P</math> be the amount deposited, <math>t</math> the number of years, <math>A</math> the balance,  
 
   Let <math>P</math> be the amount deposited, <math>t</math> the number of years, <math>A</math> the balance,  
 
   and <math>r</math> the annual interest rate (in decimal form).
 
   and <math>r</math> the annual interest rate (in decimal form).
−
   1. Compounded <math>n</math> times per year: <math>A=P(1+\frac{r}{n})^{rt}</math>
+
   1. Compounded <math>n</math> times per year: <math>A=P(1+\frac{r}{n})^{nt}</math>
 
   2. Compounded continuously: <math>A=Pe^{rt}</math>
 
   2. Compounded continuously: <math>A=Pe^{rt}</math>
  
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!Solution: &nbsp;
 
!Solution: &nbsp;
 
|-
 
|-
−
|<math>(8^{\frac{1}{2}})(2^{\frac{1}{2}})=(8\cdot 2)^{\frac{1}{2}}=16^{\frac{1}{2}}</math>
+
|<math>A=3000(1+\frac{0.04}{4})^{(4)10}</math>
 
|}
 
|}
  
−
'''a)''' Quarterly
+
'''a)''' Annually
 
{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Solution: &nbsp;
 
!Solution: &nbsp;
 
|-
 
|-
−
|<math>(8^{\frac{1}{2}})(2^{\frac{1}{2}})=(8\cdot 2)^{\frac{1}{2}}=16^{\frac{1}{2}}</math>
+
|<math>A=3000(1+\frac{0.04}{1})^{(1)10}</math>
 
|}
 
|}
  
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!Solution: &nbsp;
 
!Solution: &nbsp;
 
|-
 
|-
−
|<math>(8^{\frac{1}{2}})(2^{\frac{1}{2}})=(8\cdot 2)^{\frac{1}{2}}=16^{\frac{1}{2}}</math>
+
|<math>A=3000(1+\frac{0.04}{12})^{(12)10}</math>
 
|}
 
|}
  
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!Solution: &nbsp;
 
!Solution: &nbsp;
 
|-
 
|-
−
|<math>(8^{\frac{1}{2}})(2^{\frac{1}{2}})=(8\cdot 2)^{\frac{1}{2}}=16^{\frac{1}{2}}</math>
+
|<math>A=3000(1+\frac{0.04}{365})^{(365)10}</math>
 
|}
 
|}
  
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|}
 
|}
  
 +
==Present Value==
 +
 +
<math>P=\frac{A}{(1+\frac{r}{n})^{n t}}</math>
 +
 +
'''Exercises''' How much money should be deposited in an account paying 5% interest compounded monthly in order to have a balance of $20000 after 5 years?
 +
 +
{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 +
!Solution: &nbsp;
 +
|-
 +
|This is present value problem. So
 +
|-
 +
|<math>P=\frac{A}{(1+\frac{r}{n})^{n t}}=\frac{20000}{(1+\frac{0.05}{12})^{(12)(5)}}</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]]'''

Latest revision as of 07:12, 11 August 2020

Limit Definition of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e} 

 The irrational number Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e}
 is defined to be the limit:
 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\to 0} (1+x)^{\frac{1}{x}}=e}

 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e\approx 2.71828182846}

Compound Interest

 Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P}
 be the amount deposited, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t}
 the number of years, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A}
 the balance, 
 and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r}
 the annual interest rate (in decimal form).
 1. Compounded Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n}
 times per year: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A=P(1+\frac{r}{n})^{nt}}

 2. Compounded continuously: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A=Pe^{rt}}

Exercises Find the balance in an account when $3000 is deposited for 10 years at an interest rate of 4%, compounded as follows.

a) Quarterly

Solution:  
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A=3000(1+\frac{0.04}{4})^{(4)10}}

a) Annually

Solution:  
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A=3000(1+\frac{0.04}{1})^{(1)10}}

a) Monthly

Solution:  
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A=3000(1+\frac{0.04}{12})^{(12)10}}

a) Daily

Solution:  
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A=3000(1+\frac{0.04}{365})^{(365)10}}

a) Continuously

Solution:  
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A=Pe^{rt}=3000(e^{(0.04) (10)})}

Present Value

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P=\frac{A}{(1+\frac{r}{n})^{n t}}}

Exercises How much money should be deposited in an account paying 5% interest compounded monthly in order to have a balance of $20000 after 5 years?

Solution:  
This is present value problem. So
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P=\frac{A}{(1+\frac{r}{n})^{n t}}=\frac{20000}{(1+\frac{0.05}{12})^{(12)(5)}}}

Return to Topics Page

This page were made by Tri Phan

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