Why below question is considered as Linear and not exponential growth?
$begingroup$
A ball falls from a height of $2$ meters onto a firm surface and jumps
after each impact each back to $80%$ of the height from which it fell. ¨
Set up the function, which indicates the height of the ball after the
$n$th impact reached. How high does the ball jump after the $5$th impact?
Below question is considered as a linear growth and not exponential growth .
I dont understand why its linear growth or decay.
$$y=2*0.8^5$$
For the percentage neither it add to $1$ nor minus from $1$.
I wish to ask why it's not $y=2*.2^5$ (I took $80%$ as decay and did $1-.8=.2$)
calculus linear-algebra exponential-function
asked Dec 15 '18 at 12:02
tomtomtomtom
63
$endgroup$
add a comment |
$begingroup$
A ball falls from a height of $2$ meters onto a firm surface and jumps
after each impact each back to $80%$ of the height from which it fell. ¨
Set up the function, which indicates the height of the ball after the
$n$th impact reached. How high does the ball jump after the $5$th impact?
Below question is considered as a linear growth and not exponential growth .
I dont understand why its linear growth or decay.
$$y=2*0.8^5$$
For the percentage neither it add to $1$ nor minus from $1$.
I wish to ask why it's not $y=2*.2^5$ (I took $80%$ as decay and did $1-.8=.2$)
calculus linear-algebra exponential-function
asked Dec 15 '18 at 12:02
tomtomtomtom
63
$endgroup$
add a comment |
$begingroup$
A ball falls from a height of $2$ meters onto a firm surface and jumps
after each impact each back to $80%$ of the height from which it fell. ¨
Set up the function, which indicates the height of the ball after the
$n$th impact reached. How high does the ball jump after the $5$th impact?
Below question is considered as a linear growth and not exponential growth .
I dont understand why its linear growth or decay.
$$y=2*0.8^5$$
For the percentage neither it add to $1$ nor minus from $1$.
I wish to ask why it's not $y=2*.2^5$ (I took $80%$ as decay and did $1-.8=.2$)
calculus linear-algebra exponential-function
asked Dec 15 '18 at 12:02
tomtomtomtom
63
$endgroup$
A ball falls from a height of $2$ meters onto a firm surface and jumps
after each impact each back to $80%$ of the height from which it fell. ¨
Set up the function, which indicates the height of the ball after the
$n$th impact reached. How high does the ball jump after the $5$th impact?
Below question is considered as a linear growth and not exponential growth .
I dont understand why its linear growth or decay.
$$y=2*0.8^5$$
For the percentage neither it add to $1$ nor minus from $1$.
I wish to ask why it's not $y=2*.2^5$ (I took $80%$ as decay and did $1-.8=.2$)
calculus linear-algebra exponential-function
calculus linear-algebra exponential-function
asked Dec 15 '18 at 12:02
tomtomtomtom
63
asked Dec 15 '18 at 12:02
tomtomtomtom
63
edited Dec 16 '18 at 8:55
tomtom
asked Dec 15 '18 at 12:02
tomtomtomtom
63
asked Dec 15 '18 at 12:02
tomtomtomtom
63
asked Dec 15 '18 at 12:02
tomtomtomtom
63
63
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here's what you've done.
With each impact, the height of the bounce decreases to $80%$ of what it was. But you've made it decrease by $80%$ (to $20%$) by subtracting the $80%$ from $1$.
I think that's where your confusion is.
You need to multiply the height by $80%$ with each bounce, and the decay comes from the fact that $80%$ is smaller than $1$.
Now, looking at it step by step:
After no bounces (the starting position), you've multiplied by $0.8$ no times, so the height in metres is $$y=2=2*0.8^0$$
($0.8^0$ simply means "Don't multiply by $0.8$ at all": anything to the power of $0$ is $1$. But putting it in helps to show the pattern.)
After $1$ bounce, you've multiplied by $0.8$ once, so $$y=2*0.8=2*0.8^1$$
After $2$ bounces, you've multiplied by $0.8$ twice, so now $$y=2*0.8*0.8=2*0.8^2$$
After $3$ bounces, you've multiplied by $0.8$ $3$ times, making $$y=2*0.8*0.8*0.8=2*0.8^3$$
After $n$ bounces, you've multiplied by $0.8$ $n$ times, so $$y=2*0.8^n$$ which is the equation the question wants you to use.
Then just put $n=5$ to get the height after $5$ bounces.
Edit: When a problem like this is described in words, "of" is very often the same as "multiplied by": e.g. "two thirds of $x$" means $frac23*x$ and "$80%$ of $y$" means $80%*y$.
answered Dec 15 '18 at 12:48
timtfjtimtfj
2,238420
$endgroup$
$begingroup$
My question is why its linear growth and not an exponential growth
$endgroup$
– tomtom
Dec 16 '18 at 8:56
$begingroup$
It's not linear—that would mean adding or subtracting a fixed amount with each bounce. It's exponential because we're multiplying by a fixed amount ($0.8$) each time, and it's decay because repeatedly multiplying something by $0.8$ makes it get smaller and smaller. If we were multiplying by something bigger than $1$ (e. g. $1.1$ in $y=a*1.1^n$) then it would be growth.
$endgroup$
– timtfj
Dec 16 '18 at 11:40
$begingroup$
i am quite lost . eg the value of a new car in 2015 is 4000. it depreciates 7% every year. how much will the car be worth in 2024. here i notice that solution is first find (1-0.07)= 0.93 and then 4000(0.93)^9 is the answer. may i ask why some time for decay we minus decaly percentage from 1 and in above we just dont . may i ask why
$endgroup$
– tomtom
Dec 16 '18 at 20:03
$begingroup$
If it depreciates by $7%$, it goes down to $93%$ of what it was. So, multiply by $0.93$, $9$ times (for the $9$ years): $4000*0.93^9=2081.64$
$endgroup$
– timtfj
Dec 16 '18 at 20:13
$begingroup$
It's the difference between decreasing by a certain amount (which needs subtracting from $100%$) and decreasing to a certain amount. The key thing is to work out what multiple the new value is of the old one.
$endgroup$
– timtfj
Dec 16 '18 at 20:17
|
show 1 more comment
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$begingroup$
Here's what you've done.
With each impact, the height of the bounce decreases to $80%$ of what it was. But you've made it decrease by $80%$ (to $20%$) by subtracting the $80%$ from $1$.
I think that's where your confusion is.
You need to multiply the height by $80%$ with each bounce, and the decay comes from the fact that $80%$ is smaller than $1$.
Now, looking at it step by step:
After no bounces (the starting position), you've multiplied by $0.8$ no times, so the height in metres is $$y=2=2*0.8^0$$
($0.8^0$ simply means "Don't multiply by $0.8$ at all": anything to the power of $0$ is $1$. But putting it in helps to show the pattern.)
After $1$ bounce, you've multiplied by $0.8$ once, so $$y=2*0.8=2*0.8^1$$
After $2$ bounces, you've multiplied by $0.8$ twice, so now $$y=2*0.8*0.8=2*0.8^2$$
After $3$ bounces, you've multiplied by $0.8$ $3$ times, making $$y=2*0.8*0.8*0.8=2*0.8^3$$
After $n$ bounces, you've multiplied by $0.8$ $n$ times, so $$y=2*0.8^n$$ which is the equation the question wants you to use.
Then just put $n=5$ to get the height after $5$ bounces.
Edit: When a problem like this is described in words, "of" is very often the same as "multiplied by": e.g. "two thirds of $x$" means $frac23*x$ and "$80%$ of $y$" means $80%*y$.
answered Dec 15 '18 at 12:48
timtfjtimtfj
2,238420
$endgroup$
$begingroup$
My question is why its linear growth and not an exponential growth
$endgroup$
– tomtom
Dec 16 '18 at 8:56
$begingroup$
It's not linear—that would mean adding or subtracting a fixed amount with each bounce. It's exponential because we're multiplying by a fixed amount ($0.8$) each time, and it's decay because repeatedly multiplying something by $0.8$ makes it get smaller and smaller. If we were multiplying by something bigger than $1$ (e. g. $1.1$ in $y=a*1.1^n$) then it would be growth.
$endgroup$
– timtfj
Dec 16 '18 at 11:40
$begingroup$
i am quite lost . eg the value of a new car in 2015 is 4000. it depreciates 7% every year. how much will the car be worth in 2024. here i notice that solution is first find (1-0.07)= 0.93 and then 4000(0.93)^9 is the answer. may i ask why some time for decay we minus decaly percentage from 1 and in above we just dont . may i ask why
$endgroup$
– tomtom
Dec 16 '18 at 20:03
$begingroup$
If it depreciates by $7%$, it goes down to $93%$ of what it was. So, multiply by $0.93$, $9$ times (for the $9$ years): $4000*0.93^9=2081.64$
$endgroup$
– timtfj
Dec 16 '18 at 20:13
$begingroup$
It's the difference between decreasing by a certain amount (which needs subtracting from $100%$) and decreasing to a certain amount. The key thing is to work out what multiple the new value is of the old one.
$endgroup$
– timtfj
Dec 16 '18 at 20:17
|
show 1 more comment
$begingroup$
Here's what you've done.
With each impact, the height of the bounce decreases to $80%$ of what it was. But you've made it decrease by $80%$ (to $20%$) by subtracting the $80%$ from $1$.
I think that's where your confusion is.
You need to multiply the height by $80%$ with each bounce, and the decay comes from the fact that $80%$ is smaller than $1$.
Now, looking at it step by step:
After no bounces (the starting position), you've multiplied by $0.8$ no times, so the height in metres is $$y=2=2*0.8^0$$
($0.8^0$ simply means "Don't multiply by $0.8$ at all": anything to the power of $0$ is $1$. But putting it in helps to show the pattern.)
After $1$ bounce, you've multiplied by $0.8$ once, so $$y=2*0.8=2*0.8^1$$
After $2$ bounces, you've multiplied by $0.8$ twice, so now $$y=2*0.8*0.8=2*0.8^2$$
After $3$ bounces, you've multiplied by $0.8$ $3$ times, making $$y=2*0.8*0.8*0.8=2*0.8^3$$
After $n$ bounces, you've multiplied by $0.8$ $n$ times, so $$y=2*0.8^n$$ which is the equation the question wants you to use.
Then just put $n=5$ to get the height after $5$ bounces.
Edit: When a problem like this is described in words, "of" is very often the same as "multiplied by": e.g. "two thirds of $x$" means $frac23*x$ and "$80%$ of $y$" means $80%*y$.
answered Dec 15 '18 at 12:48
timtfjtimtfj
2,238420
$endgroup$
$begingroup$
My question is why its linear growth and not an exponential growth
$endgroup$
– tomtom
Dec 16 '18 at 8:56
$begingroup$
It's not linear—that would mean adding or subtracting a fixed amount with each bounce. It's exponential because we're multiplying by a fixed amount ($0.8$) each time, and it's decay because repeatedly multiplying something by $0.8$ makes it get smaller and smaller. If we were multiplying by something bigger than $1$ (e. g. $1.1$ in $y=a*1.1^n$) then it would be growth.
$endgroup$
– timtfj
Dec 16 '18 at 11:40
$begingroup$
i am quite lost . eg the value of a new car in 2015 is 4000. it depreciates 7% every year. how much will the car be worth in 2024. here i notice that solution is first find (1-0.07)= 0.93 and then 4000(0.93)^9 is the answer. may i ask why some time for decay we minus decaly percentage from 1 and in above we just dont . may i ask why
$endgroup$
– tomtom
Dec 16 '18 at 20:03
$begingroup$
If it depreciates by $7%$, it goes down to $93%$ of what it was. So, multiply by $0.93$, $9$ times (for the $9$ years): $4000*0.93^9=2081.64$
$endgroup$
– timtfj
Dec 16 '18 at 20:13
$begingroup$
It's the difference between decreasing by a certain amount (which needs subtracting from $100%$) and decreasing to a certain amount. The key thing is to work out what multiple the new value is of the old one.
$endgroup$
– timtfj
Dec 16 '18 at 20:17
|
show 1 more comment
$begingroup$
Here's what you've done.
With each impact, the height of the bounce decreases to $80%$ of what it was. But you've made it decrease by $80%$ (to $20%$) by subtracting the $80%$ from $1$.
I think that's where your confusion is.
You need to multiply the height by $80%$ with each bounce, and the decay comes from the fact that $80%$ is smaller than $1$.
Now, looking at it step by step:
After no bounces (the starting position), you've multiplied by $0.8$ no times, so the height in metres is $$y=2=2*0.8^0$$
($0.8^0$ simply means "Don't multiply by $0.8$ at all": anything to the power of $0$ is $1$. But putting it in helps to show the pattern.)
After $1$ bounce, you've multiplied by $0.8$ once, so $$y=2*0.8=2*0.8^1$$
After $2$ bounces, you've multiplied by $0.8$ twice, so now $$y=2*0.8*0.8=2*0.8^2$$
After $3$ bounces, you've multiplied by $0.8$ $3$ times, making $$y=2*0.8*0.8*0.8=2*0.8^3$$
After $n$ bounces, you've multiplied by $0.8$ $n$ times, so $$y=2*0.8^n$$ which is the equation the question wants you to use.
Then just put $n=5$ to get the height after $5$ bounces.
Edit: When a problem like this is described in words, "of" is very often the same as "multiplied by": e.g. "two thirds of $x$" means $frac23*x$ and "$80%$ of $y$" means $80%*y$.
answered Dec 15 '18 at 12:48
timtfjtimtfj
2,238420
$endgroup$
Here's what you've done.
With each impact, the height of the bounce decreases to $80%$ of what it was. But you've made it decrease by $80%$ (to $20%$) by subtracting the $80%$ from $1$.
I think that's where your confusion is.
You need to multiply the height by $80%$ with each bounce, and the decay comes from the fact that $80%$ is smaller than $1$.
Now, looking at it step by step:
After no bounces (the starting position), you've multiplied by $0.8$ no times, so the height in metres is $$y=2=2*0.8^0$$
($0.8^0$ simply means "Don't multiply by $0.8$ at all": anything to the power of $0$ is $1$. But putting it in helps to show the pattern.)
After $1$ bounce, you've multiplied by $0.8$ once, so $$y=2*0.8=2*0.8^1$$
After $2$ bounces, you've multiplied by $0.8$ twice, so now $$y=2*0.8*0.8=2*0.8^2$$
After $3$ bounces, you've multiplied by $0.8$ $3$ times, making $$y=2*0.8*0.8*0.8=2*0.8^3$$
After $n$ bounces, you've multiplied by $0.8$ $n$ times, so $$y=2*0.8^n$$ which is the equation the question wants you to use.
Then just put $n=5$ to get the height after $5$ bounces.
Edit: When a problem like this is described in words, "of" is very often the same as "multiplied by": e.g. "two thirds of $x$" means $frac23*x$ and "$80%$ of $y$" means $80%*y$.
answered Dec 15 '18 at 12:48
timtfjtimtfj
2,238420
edited Dec 15 '18 at 16:54
answered Dec 15 '18 at 12:48
timtfjtimtfj
2,238420
answered Dec 15 '18 at 12:48
timtfjtimtfj
2,238420
answered Dec 15 '18 at 12:48
timtfjtimtfj
2,238420
2,238420
$begingroup$
My question is why its linear growth and not an exponential growth
$endgroup$
– tomtom
Dec 16 '18 at 8:56
$begingroup$
It's not linear—that would mean adding or subtracting a fixed amount with each bounce. It's exponential because we're multiplying by a fixed amount ($0.8$) each time, and it's decay because repeatedly multiplying something by $0.8$ makes it get smaller and smaller. If we were multiplying by something bigger than $1$ (e. g. $1.1$ in $y=a*1.1^n$) then it would be growth.
$endgroup$
– timtfj
Dec 16 '18 at 11:40
$begingroup$
i am quite lost . eg the value of a new car in 2015 is 4000. it depreciates 7% every year. how much will the car be worth in 2024. here i notice that solution is first find (1-0.07)= 0.93 and then 4000(0.93)^9 is the answer. may i ask why some time for decay we minus decaly percentage from 1 and in above we just dont . may i ask why
$endgroup$
– tomtom
Dec 16 '18 at 20:03
$begingroup$
If it depreciates by $7%$, it goes down to $93%$ of what it was. So, multiply by $0.93$, $9$ times (for the $9$ years): $4000*0.93^9=2081.64$
$endgroup$
– timtfj
Dec 16 '18 at 20:13
$begingroup$
It's the difference between decreasing by a certain amount (which needs subtracting from $100%$) and decreasing to a certain amount. The key thing is to work out what multiple the new value is of the old one.
$endgroup$
– timtfj
Dec 16 '18 at 20:17
|
show 1 more comment
$begingroup$
My question is why its linear growth and not an exponential growth
$endgroup$
– tomtom
Dec 16 '18 at 8:56
$begingroup$
It's not linear—that would mean adding or subtracting a fixed amount with each bounce. It's exponential because we're multiplying by a fixed amount ($0.8$) each time, and it's decay because repeatedly multiplying something by $0.8$ makes it get smaller and smaller. If we were multiplying by something bigger than $1$ (e. g. $1.1$ in $y=a*1.1^n$) then it would be growth.
$endgroup$
– timtfj
Dec 16 '18 at 11:40
$begingroup$
i am quite lost . eg the value of a new car in 2015 is 4000. it depreciates 7% every year. how much will the car be worth in 2024. here i notice that solution is first find (1-0.07)= 0.93 and then 4000(0.93)^9 is the answer. may i ask why some time for decay we minus decaly percentage from 1 and in above we just dont . may i ask why
$endgroup$
– tomtom
Dec 16 '18 at 20:03
$begingroup$
If it depreciates by $7%$, it goes down to $93%$ of what it was. So, multiply by $0.93$, $9$ times (for the $9$ years): $4000*0.93^9=2081.64$
$endgroup$
– timtfj
Dec 16 '18 at 20:13
$begingroup$
It's the difference between decreasing by a certain amount (which needs subtracting from $100%$) and decreasing to a certain amount. The key thing is to work out what multiple the new value is of the old one.
$endgroup$
– timtfj
Dec 16 '18 at 20:17
$begingroup$
My question is why its linear growth and not an exponential growth
$endgroup$
– tomtom
Dec 16 '18 at 8:56
$begingroup$
My question is why its linear growth and not an exponential growth
$endgroup$
– tomtom
Dec 16 '18 at 8:56
$begingroup$
It's not linear—that would mean adding or subtracting a fixed amount with each bounce. It's exponential because we're multiplying by a fixed amount ($0.8$) each time, and it's decay because repeatedly multiplying something by $0.8$ makes it get smaller and smaller. If we were multiplying by something bigger than $1$ (e. g. $1.1$ in $y=a*1.1^n$) then it would be growth.
$endgroup$
– timtfj
Dec 16 '18 at 11:40
$begingroup$
It's not linear—that would mean adding or subtracting a fixed amount with each bounce. It's exponential because we're multiplying by a fixed amount ($0.8$) each time, and it's decay because repeatedly multiplying something by $0.8$ makes it get smaller and smaller. If we were multiplying by something bigger than $1$ (e. g. $1.1$ in $y=a*1.1^n$) then it would be growth.
$endgroup$
– timtfj
Dec 16 '18 at 11:40
$begingroup$
i am quite lost . eg the value of a new car in 2015 is 4000. it depreciates 7% every year. how much will the car be worth in 2024. here i notice that solution is first find (1-0.07)= 0.93 and then 4000(0.93)^9 is the answer. may i ask why some time for decay we minus decaly percentage from 1 and in above we just dont . may i ask why
$endgroup$
– tomtom
Dec 16 '18 at 20:03
$begingroup$
i am quite lost . eg the value of a new car in 2015 is 4000. it depreciates 7% every year. how much will the car be worth in 2024. here i notice that solution is first find (1-0.07)= 0.93 and then 4000(0.93)^9 is the answer. may i ask why some time for decay we minus decaly percentage from 1 and in above we just dont . may i ask why
$endgroup$
– tomtom
Dec 16 '18 at 20:03
$begingroup$
If it depreciates by $7%$, it goes down to $93%$ of what it was. So, multiply by $0.93$, $9$ times (for the $9$ years): $4000*0.93^9=2081.64$
$endgroup$
– timtfj
Dec 16 '18 at 20:13
$begingroup$
If it depreciates by $7%$, it goes down to $93%$ of what it was. So, multiply by $0.93$, $9$ times (for the $9$ years): $4000*0.93^9=2081.64$
$endgroup$
– timtfj
Dec 16 '18 at 20:13
$begingroup$
It's the difference between decreasing by a certain amount (which needs subtracting from $100%$) and decreasing to a certain amount. The key thing is to work out what multiple the new value is of the old one.
$endgroup$
– timtfj
Dec 16 '18 at 20:17
$begingroup$
It's the difference between decreasing by a certain amount (which needs subtracting from $100%$) and decreasing to a certain amount. The key thing is to work out what multiple the new value is of the old one.
$endgroup$
– timtfj
Dec 16 '18 at 20:17
|
show 1 more comment
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Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
