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This is one of our many quizzes on Exponential growth and decay. Click on this to access more such competitive quizzes on Differential Calculus.

Exponential growth and decay Quiz 2

mathematics

 Which of the following equations shows the correct equation for the following given diagram.<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/b59fddf4-606e-4be4-b89a-eef54bc8c4093912512713521761904.png"/>

 Use the formula $y\left( x \right)=a{{e}^{-ux}}$. Then substitute $x = 8$ and to each option and record value of . If the calculated ordered pair  matches to the given graph then that will be the right option.

  $\text{In the given figure, take a point }x = 8\text{ as a guide and substitute in given equations.} \\ y\left( 8 \right)=13{{e}^{-0.54\times 8}}=13{{e}^{-4.32}}=13\times 0.013=0.169. \\ \text{Since, only the point }(8, 0.169)\text{ corresponds to the following figure.} \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/71579293-c0a7-4f4e-8dc0-3d9210f9c4fd6286588099236952085.png"/>

 Find out the correct value of a if the equation given to us is $y\left( x \right)=a{{e}^{-125x}}$ along with the following diagram, <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/f49f8575-0c38-4577-a82e-938e21bdefb03034455157576802201.png"/>

 Use the formula $y\left( x \right)=a{{e}^{-ux}}$ and consider a point from y-axis and substitute in each option and record value of $x$. If the calculated ordered pair $(x, y)$ matches to the given graph then that will be the right option.

  $\text{Consider }y = 0.05\text{ and substitute this point in all four options.} \\ \text{Now, option (c) as, } \\ y\left( x \right)=45{{e}^{-125x}} \\ \Rightarrow 0.05=45{{e}^{-125x}} \\ \Rightarrow {{e}^{-125x}}=\dfrac{0.05}{45}=0.0011 \\ \Rightarrow \ln \left( {{e}^{-125x}} \right)=\ln \left( 0.0011 \right) \\ \Rightarrow -125x=-68.12 \\ \Rightarrow x=\dfrac{68.12}{125}=0.544 \\ \Rightarrow \left( x,y \right)=\left( 0.544,0.05 \right) \\ \text{The only point that matches to the following fig is } (0.0544, 0.05). \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/d48c168d-6abe-4142-a8ae-60ab3ffa69f3225487375741411497.png"/>

 What if the population of bacteria is 500 at first and is exponentially decreasing. After 3 years the population reaches 150. So, how much time will it take to reach - 25?

 Use the formula $y\left( x \right)=a{{e}^{-ux}}$, where $a$ is 500 in the population and $x$ is 3 years. After this substitution, apply logarithm to this equation.

  $\text{The bacteria are decreasing exponentially, } \\ y\left( x \right)=a{{e}^{-ux}}  -----(i) \\ \text{The bacteria’s population was } 500 \text{ at first so, substitute } x = 0 \text{ in (i), which gives }y\left( 0 \right)=500=a{{e}^{-u0}}=a{{e}^{0}}=a\left[ \because {{e}^{0}}=1 \right] \\ a=500  -----(ii) \\ \text{After 3 years, }y\left( 3 \right)=150=a{{e}^{-u3}}=a{{e}^{-3u}} \\ \text{From equation (ii),} \\ \Rightarrow {{e}^{-3u}}=\dfrac{150}{500} \\ \ln \left( {{e}^{-3u}} \right)=\ln \left( 0.3 \right) \\ \Rightarrow -3u=-1.20\left[ \text{Since},\,\ln {{e}^{t}}=t\,\text{and}\,\ln \left( 0.3 \right)=-1.20 \right] \\ \Rightarrow u=0.4\,...(iii) \\ \text{Substitute (ii) and (iii) in equation (i),} \\ y\left( x \right)=500{{e}^{-0.4x}} \\ \text{This equation shows the following exponential graph.} \\ \text{Now, for the bacteria to reach }-25, \\ 500{{e}^{-0.4x}}=-25 \\ \Rightarrow {{e}^{-0.4x}}=-\dfrac{25}{500} \\ \Rightarrow \ln \left( {{e}^{-0.4x}} \right)=-0.05 \\ \Rightarrow x=0.125 \\ $ So, it will take 0.125 years to reach the negative of 25 in terms of population.$ \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/addd7c85-5265-4f4e-a68d-289720db545c8728776662368890548.png"/>

 Suppose that a bacteria $X$ is decaying into $Y$. At first the scientists noticed the total weight of $X$ as 64 milligrams. After 10 years, will it vanish? $\text{(Given the half life of bacteria ‘X’ as 6421 years)}$

 The question is about exponentially decreasing, use the formula $y\left( x \right)=a{{e}^{-ux}}$. Then use the formula $\text{Half life}=\dfrac{\ln 2}{u}$. After this apply logarithm to the equation.

  $\text{The bacteria are decreasing exponentially, } \\ y\left( x \right)=a{{e}^{-ux}}  ---(i) \\ \text{At starting decaying will be at 0. } \\ y\left( 0 \right)=64=a{{e}^{-u0}}=a{{e}^{0}}=a\left[ \because {{e}^{0}}=1 \right] \\ a=64  -----(ii) \\ \text{Half life}=\dfrac{\ln 2}{u} \\ 6=\dfrac{\ln 2}{u} \\ \Rightarrow u=\dfrac{0.69}{6} \\ \Rightarrow u=0.115...(iii) \\ \text{Substituting (ii) and (iii) in (i) } \\ y\left( x \right)=64{{e}^{-0.115x}} \\ \text{As time is }10\text{years, }x = 10\text{ gives, } \\ y\left( 10 \right)=64{{e}^{-0.115\times 10}} \\ \Rightarrow y\left( 10 \right)=64\times 0.316\left[ \text{Since, }{{e}^{-1.15}}=0.316 \right] \\ \Rightarrow y\left( 10 \right)=20.224 \\ $ So, only 20.224 milligrams of the bacteria  $X$ will be left after the decay for 10 years $ \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/d1f059ac-5825-4f48-b4bf-a39354e926b42031089238489356417.png"/>

 A housewife brings boiled water, at a temperature of 140 F. The actual temperature just fits for drinking is 96 Fahrenheit for her daughter. The temperature in the winter season was 26 Fahrenheit. Left the glass of water for 1 minute, the temperature gets reduced to 110 Fahrenheit. Was it the right time to drink it?

 Use $T=\left( {{T}_{o}}-{{T}_{a}} \right){{e}^{kt}}+{{T}_{a}}$ where ${{T}_{o}}=140$ as the initial temperature and ${{T}_{a}}=26$ as the temperature of the room. Also, apply a log function.

  $T=\left( {{T}_{o}}-{{T}_{a}} \right){{e}^{kt}}+{{T}_{a}} ----(i) \\ \text{Take }{{T}_{o}}=140\text{ as the initial temperature and } {{T}_{a}}=26 \text{ as the temperature of the room. } \\ \text{Substitute }T = 110\text{ in equation (i),} \\ T=\left( {{T}_{o}}-{{T}_{a}} \right){{e}^{kt}}+{{T}_{a}} \\ \Rightarrow 110=\left( 140-26 \right){{e}^{kt}}+26 \\ \Rightarrow 110=114{{e}^{kt}}+26 \\ \text{Since, time was } 1 \text{ minute when temp decreases to } 110F\text{ so,} \\ 110=114{{e}^{k\left( 1 \right)}}+26 \\ \Rightarrow 84=114{{e}^{k}} \\ \Rightarrow {{e}^{k}}=\dfrac{84}{114}=0.73 \\ \Rightarrow \ln \left( {{e}^{k}} \right)=\ln \left( 0.73 \right) \\ \Rightarrow k=-0.314....(ii) \\ \text{Since, the right temp for her daughter was 96F} \\ \therefore T=\left( {{T}_{o}}-{{T}_{a}} \right){{e}^{kt}}+{{T}_{a}}\text{ changes to, } \\ 96=\left( 140-26 \right){{e}^{kt}}+26 \\ \Rightarrow 96-26=114{{e}^{-0.314t}}\left[ \text{Since, by (ii)} \right] \\ \Rightarrow 70=114{{e}^{-0.314t}} \\ \Rightarrow \ln \left( {{e}^{-0.314t}} \right)=\ln \left( 0.614 \right) \\ \Rightarrow t=\dfrac{0.487}{0.314}=1.55 \\ \text{thus, right time to drink water here is after }1.55min.$