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Calculus Made Easy Exercise 2 Question 12 (Converting length to diameter?)
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In Calculus Made Easy by Silvanus P. Thompson, there's a question I do not understand the logic of. Specifically, I don't understand how I should find the rate of variation of a diameter ($D$) when he only gives the length ($L$) (original question below). I've searched for a step-by-step solution for this problem, but could not find one. I'm sure I'm missing something obvious, but I just don't know what it is.
Problem: The length $L$ of an iron rod at the temperature $T$ begin given by $L=l_t [1+0.000012(T-t)]$, where $l_t$ is the length at temperature $t$, find the rate of variation of the diameter $D$ of an iron tyre suitable for being shrunk on a wheel, when the temperature $T$ varies.
calculus
asked Jul 19 at 17:33
marcski55
33
add a comment |Â
up vote
0
down vote
favorite
In Calculus Made Easy by Silvanus P. Thompson, there's a question I do not understand the logic of. Specifically, I don't understand how I should find the rate of variation of a diameter ($D$) when he only gives the length ($L$) (original question below). I've searched for a step-by-step solution for this problem, but could not find one. I'm sure I'm missing something obvious, but I just don't know what it is.
Problem: The length $L$ of an iron rod at the temperature $T$ begin given by $L=l_t [1+0.000012(T-t)]$, where $l_t$ is the length at temperature $t$, find the rate of variation of the diameter $D$ of an iron tyre suitable for being shrunk on a wheel, when the temperature $T$ varies.
calculus
asked Jul 19 at 17:33
marcski55
33
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In Calculus Made Easy by Silvanus P. Thompson, there's a question I do not understand the logic of. Specifically, I don't understand how I should find the rate of variation of a diameter ($D$) when he only gives the length ($L$) (original question below). I've searched for a step-by-step solution for this problem, but could not find one. I'm sure I'm missing something obvious, but I just don't know what it is.
Problem: The length $L$ of an iron rod at the temperature $T$ begin given by $L=l_t [1+0.000012(T-t)]$, where $l_t$ is the length at temperature $t$, find the rate of variation of the diameter $D$ of an iron tyre suitable for being shrunk on a wheel, when the temperature $T$ varies.
calculus
asked Jul 19 at 17:33
marcski55
33
In Calculus Made Easy by Silvanus P. Thompson, there's a question I do not understand the logic of. Specifically, I don't understand how I should find the rate of variation of a diameter ($D$) when he only gives the length ($L$) (original question below). I've searched for a step-by-step solution for this problem, but could not find one. I'm sure I'm missing something obvious, but I just don't know what it is.
Problem: The length $L$ of an iron rod at the temperature $T$ begin given by $L=l_t [1+0.000012(T-t)]$, where $l_t$ is the length at temperature $t$, find the rate of variation of the diameter $D$ of an iron tyre suitable for being shrunk on a wheel, when the temperature $T$ varies.
calculus
asked Jul 19 at 17:33
marcski55
33
asked Jul 19 at 17:33
marcski55
33
asked Jul 19 at 17:33
marcski55
33
asked Jul 19 at 17:33
marcski55
33
33
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
The idea is that you imagine bending the rod, at temperature $t$, of length $L = ell_t$ into a circle (whose diameter is therefore $fracell_tpi$). The diameter of this circular tyre at temperature $T$ is then given by
beginalign
D
&= fracLpi\
&= fracell_t left[1 + 0.000012(T-t) right]pi\
&= ell_t left[frac1pi + frac0.000012pi(T-t) right]\
&approx ell_t left[0.3183 + .00000382(T-t) right]\
endalign
which is a function of the (varying) temperature $T$.
Post-comment addition
The number $k = 0.000012$ is the "(temperature) coefficient of expansion" of iron. (I have no idea whether the value is correct or not, but that's beside the point). The idea is that for each degree of temperature change, a piece of metal of size $s$ will increase its size in proportion to $s$, with the constant of proportionality being $k$. So a 200-foot long piece of metal will increase in size by $200k = .000012 times 200 = .0024$ feet.
Now let's look at that iron rod, and suppose, to make the arithmetic work out nicely, that it's $3.14 approx pi$ feet long. Then when we bend it into a circle, the circle's diameter will be $1$ foot.
If we raise the temperature of the rod by 1 degree, it'll grow by $3.14 times 0.000012$ feet. That is to say, the circumference of the circle will grow by that amount.
What about the circle's diameter? Well, we could replace the circle with a circle drawn on a piece of sheet metal, and clearly the diameter of that sheet-metal circle would increase by a factor of $0.000012$; since the circle has diameter $1$ foot, the actual increase will be $0.000012$ feet.
That is to say: when the temperature is raised by one degree, the circumference grows by $k * 3.14$ feet; the diameter by $k$ feet. That's exactly what Ross's comment says, but in disguise: the proportional growth in diameter will be $.000012$, the same as circumference. But since the circumference and diameter start out as different numbers, the actual growth in the two will differ...by a factor of $pi$.
answered Jul 19 at 17:43
John Hughes
59.4k23785
This is a badly stated question that assumes incorrect behavior on the part of the tire. I am sure you have the intended solution. In fact the diameter of the tire will expand with the linear expansion of iron, not $frac 1pi$ of it. Holes in objects expand with temperature just like they were filled-imagine a solid disk of iron. You would expect the diameter to expand with the coefficient of iron.
– Ross Millikan
Jul 19 at 17:47
@RossMillikan, this is what I thought (and originally just tried to do $fracdLdt$), but I could tell from the solution that something else was to be considered. Thank you, John Hughes, for clarifying the author's original intent of the problem as that is what was hanging me up.
– marcski55
Jul 19 at 17:54
I think you're somewhat mistaken, Ross, or at least that your comment is possibly misleading; I'll add something to my answer to clarify.
– John Hughes
Jul 19 at 18:41
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The idea is that you imagine bending the rod, at temperature $t$, of length $L = ell_t$ into a circle (whose diameter is therefore $fracell_tpi$). The diameter of this circular tyre at temperature $T$ is then given by
beginalign
D
&= fracLpi\
&= fracell_t left[1 + 0.000012(T-t) right]pi\
&= ell_t left[frac1pi + frac0.000012pi(T-t) right]\
&approx ell_t left[0.3183 + .00000382(T-t) right]\
endalign
which is a function of the (varying) temperature $T$.
Post-comment addition
The number $k = 0.000012$ is the "(temperature) coefficient of expansion" of iron. (I have no idea whether the value is correct or not, but that's beside the point). The idea is that for each degree of temperature change, a piece of metal of size $s$ will increase its size in proportion to $s$, with the constant of proportionality being $k$. So a 200-foot long piece of metal will increase in size by $200k = .000012 times 200 = .0024$ feet.
Now let's look at that iron rod, and suppose, to make the arithmetic work out nicely, that it's $3.14 approx pi$ feet long. Then when we bend it into a circle, the circle's diameter will be $1$ foot.
If we raise the temperature of the rod by 1 degree, it'll grow by $3.14 times 0.000012$ feet. That is to say, the circumference of the circle will grow by that amount.
What about the circle's diameter? Well, we could replace the circle with a circle drawn on a piece of sheet metal, and clearly the diameter of that sheet-metal circle would increase by a factor of $0.000012$; since the circle has diameter $1$ foot, the actual increase will be $0.000012$ feet.
That is to say: when the temperature is raised by one degree, the circumference grows by $k * 3.14$ feet; the diameter by $k$ feet. That's exactly what Ross's comment says, but in disguise: the proportional growth in diameter will be $.000012$, the same as circumference. But since the circumference and diameter start out as different numbers, the actual growth in the two will differ...by a factor of $pi$.
answered Jul 19 at 17:43
John Hughes
59.4k23785
This is a badly stated question that assumes incorrect behavior on the part of the tire. I am sure you have the intended solution. In fact the diameter of the tire will expand with the linear expansion of iron, not $frac 1pi$ of it. Holes in objects expand with temperature just like they were filled-imagine a solid disk of iron. You would expect the diameter to expand with the coefficient of iron.
– Ross Millikan
Jul 19 at 17:47
@RossMillikan, this is what I thought (and originally just tried to do $fracdLdt$), but I could tell from the solution that something else was to be considered. Thank you, John Hughes, for clarifying the author's original intent of the problem as that is what was hanging me up.
– marcski55
Jul 19 at 17:54
I think you're somewhat mistaken, Ross, or at least that your comment is possibly misleading; I'll add something to my answer to clarify.
– John Hughes
Jul 19 at 18:41
add a comment |Â
up vote
1
down vote
accepted
The idea is that you imagine bending the rod, at temperature $t$, of length $L = ell_t$ into a circle (whose diameter is therefore $fracell_tpi$). The diameter of this circular tyre at temperature $T$ is then given by
beginalign
D
&= fracLpi\
&= fracell_t left[1 + 0.000012(T-t) right]pi\
&= ell_t left[frac1pi + frac0.000012pi(T-t) right]\
&approx ell_t left[0.3183 + .00000382(T-t) right]\
endalign
which is a function of the (varying) temperature $T$.
Post-comment addition
The number $k = 0.000012$ is the "(temperature) coefficient of expansion" of iron. (I have no idea whether the value is correct or not, but that's beside the point). The idea is that for each degree of temperature change, a piece of metal of size $s$ will increase its size in proportion to $s$, with the constant of proportionality being $k$. So a 200-foot long piece of metal will increase in size by $200k = .000012 times 200 = .0024$ feet.
Now let's look at that iron rod, and suppose, to make the arithmetic work out nicely, that it's $3.14 approx pi$ feet long. Then when we bend it into a circle, the circle's diameter will be $1$ foot.
If we raise the temperature of the rod by 1 degree, it'll grow by $3.14 times 0.000012$ feet. That is to say, the circumference of the circle will grow by that amount.
What about the circle's diameter? Well, we could replace the circle with a circle drawn on a piece of sheet metal, and clearly the diameter of that sheet-metal circle would increase by a factor of $0.000012$; since the circle has diameter $1$ foot, the actual increase will be $0.000012$ feet.
That is to say: when the temperature is raised by one degree, the circumference grows by $k * 3.14$ feet; the diameter by $k$ feet. That's exactly what Ross's comment says, but in disguise: the proportional growth in diameter will be $.000012$, the same as circumference. But since the circumference and diameter start out as different numbers, the actual growth in the two will differ...by a factor of $pi$.
answered Jul 19 at 17:43
John Hughes
59.4k23785
This is a badly stated question that assumes incorrect behavior on the part of the tire. I am sure you have the intended solution. In fact the diameter of the tire will expand with the linear expansion of iron, not $frac 1pi$ of it. Holes in objects expand with temperature just like they were filled-imagine a solid disk of iron. You would expect the diameter to expand with the coefficient of iron.
– Ross Millikan
Jul 19 at 17:47
@RossMillikan, this is what I thought (and originally just tried to do $fracdLdt$), but I could tell from the solution that something else was to be considered. Thank you, John Hughes, for clarifying the author's original intent of the problem as that is what was hanging me up.
– marcski55
Jul 19 at 17:54
I think you're somewhat mistaken, Ross, or at least that your comment is possibly misleading; I'll add something to my answer to clarify.
– John Hughes
Jul 19 at 18:41
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The idea is that you imagine bending the rod, at temperature $t$, of length $L = ell_t$ into a circle (whose diameter is therefore $fracell_tpi$). The diameter of this circular tyre at temperature $T$ is then given by
beginalign
D
&= fracLpi\
&= fracell_t left[1 + 0.000012(T-t) right]pi\
&= ell_t left[frac1pi + frac0.000012pi(T-t) right]\
&approx ell_t left[0.3183 + .00000382(T-t) right]\
endalign
which is a function of the (varying) temperature $T$.
Post-comment addition
The number $k = 0.000012$ is the "(temperature) coefficient of expansion" of iron. (I have no idea whether the value is correct or not, but that's beside the point). The idea is that for each degree of temperature change, a piece of metal of size $s$ will increase its size in proportion to $s$, with the constant of proportionality being $k$. So a 200-foot long piece of metal will increase in size by $200k = .000012 times 200 = .0024$ feet.
Now let's look at that iron rod, and suppose, to make the arithmetic work out nicely, that it's $3.14 approx pi$ feet long. Then when we bend it into a circle, the circle's diameter will be $1$ foot.
If we raise the temperature of the rod by 1 degree, it'll grow by $3.14 times 0.000012$ feet. That is to say, the circumference of the circle will grow by that amount.
What about the circle's diameter? Well, we could replace the circle with a circle drawn on a piece of sheet metal, and clearly the diameter of that sheet-metal circle would increase by a factor of $0.000012$; since the circle has diameter $1$ foot, the actual increase will be $0.000012$ feet.
That is to say: when the temperature is raised by one degree, the circumference grows by $k * 3.14$ feet; the diameter by $k$ feet. That's exactly what Ross's comment says, but in disguise: the proportional growth in diameter will be $.000012$, the same as circumference. But since the circumference and diameter start out as different numbers, the actual growth in the two will differ...by a factor of $pi$.
answered Jul 19 at 17:43
John Hughes
59.4k23785
The idea is that you imagine bending the rod, at temperature $t$, of length $L = ell_t$ into a circle (whose diameter is therefore $fracell_tpi$). The diameter of this circular tyre at temperature $T$ is then given by
beginalign
D
&= fracLpi\
&= fracell_t left[1 + 0.000012(T-t) right]pi\
&= ell_t left[frac1pi + frac0.000012pi(T-t) right]\
&approx ell_t left[0.3183 + .00000382(T-t) right]\
endalign
which is a function of the (varying) temperature $T$.
Post-comment addition
The number $k = 0.000012$ is the "(temperature) coefficient of expansion" of iron. (I have no idea whether the value is correct or not, but that's beside the point). The idea is that for each degree of temperature change, a piece of metal of size $s$ will increase its size in proportion to $s$, with the constant of proportionality being $k$. So a 200-foot long piece of metal will increase in size by $200k = .000012 times 200 = .0024$ feet.
Now let's look at that iron rod, and suppose, to make the arithmetic work out nicely, that it's $3.14 approx pi$ feet long. Then when we bend it into a circle, the circle's diameter will be $1$ foot.
If we raise the temperature of the rod by 1 degree, it'll grow by $3.14 times 0.000012$ feet. That is to say, the circumference of the circle will grow by that amount.
What about the circle's diameter? Well, we could replace the circle with a circle drawn on a piece of sheet metal, and clearly the diameter of that sheet-metal circle would increase by a factor of $0.000012$; since the circle has diameter $1$ foot, the actual increase will be $0.000012$ feet.
That is to say: when the temperature is raised by one degree, the circumference grows by $k * 3.14$ feet; the diameter by $k$ feet. That's exactly what Ross's comment says, but in disguise: the proportional growth in diameter will be $.000012$, the same as circumference. But since the circumference and diameter start out as different numbers, the actual growth in the two will differ...by a factor of $pi$.
answered Jul 19 at 17:43
John Hughes
59.4k23785
edited Jul 19 at 18:41
answered Jul 19 at 17:43
John Hughes
59.4k23785
answered Jul 19 at 17:43
John Hughes
59.4k23785
answered Jul 19 at 17:43
John Hughes
59.4k23785
59.4k23785
This is a badly stated question that assumes incorrect behavior on the part of the tire. I am sure you have the intended solution. In fact the diameter of the tire will expand with the linear expansion of iron, not $frac 1pi$ of it. Holes in objects expand with temperature just like they were filled-imagine a solid disk of iron. You would expect the diameter to expand with the coefficient of iron.
– Ross Millikan
Jul 19 at 17:47
@RossMillikan, this is what I thought (and originally just tried to do $fracdLdt$), but I could tell from the solution that something else was to be considered. Thank you, John Hughes, for clarifying the author's original intent of the problem as that is what was hanging me up.
– marcski55
Jul 19 at 17:54
I think you're somewhat mistaken, Ross, or at least that your comment is possibly misleading; I'll add something to my answer to clarify.
– John Hughes
Jul 19 at 18:41
add a comment |Â
This is a badly stated question that assumes incorrect behavior on the part of the tire. I am sure you have the intended solution. In fact the diameter of the tire will expand with the linear expansion of iron, not $frac 1pi$ of it. Holes in objects expand with temperature just like they were filled-imagine a solid disk of iron. You would expect the diameter to expand with the coefficient of iron.
– Ross Millikan
Jul 19 at 17:47
@RossMillikan, this is what I thought (and originally just tried to do $fracdLdt$), but I could tell from the solution that something else was to be considered. Thank you, John Hughes, for clarifying the author's original intent of the problem as that is what was hanging me up.
– marcski55
Jul 19 at 17:54
I think you're somewhat mistaken, Ross, or at least that your comment is possibly misleading; I'll add something to my answer to clarify.
– John Hughes
Jul 19 at 18:41
This is a badly stated question that assumes incorrect behavior on the part of the tire. I am sure you have the intended solution. In fact the diameter of the tire will expand with the linear expansion of iron, not $frac 1pi$ of it. Holes in objects expand with temperature just like they were filled-imagine a solid disk of iron. You would expect the diameter to expand with the coefficient of iron.
– Ross Millikan
Jul 19 at 17:47
This is a badly stated question that assumes incorrect behavior on the part of the tire. I am sure you have the intended solution. In fact the diameter of the tire will expand with the linear expansion of iron, not $frac 1pi$ of it. Holes in objects expand with temperature just like they were filled-imagine a solid disk of iron. You would expect the diameter to expand with the coefficient of iron.
– Ross Millikan
Jul 19 at 17:47
@RossMillikan, this is what I thought (and originally just tried to do $fracdLdt$), but I could tell from the solution that something else was to be considered. Thank you, John Hughes, for clarifying the author's original intent of the problem as that is what was hanging me up.
– marcski55
Jul 19 at 17:54
@RossMillikan, this is what I thought (and originally just tried to do $fracdLdt$), but I could tell from the solution that something else was to be considered. Thank you, John Hughes, for clarifying the author's original intent of the problem as that is what was hanging me up.
– marcski55
Jul 19 at 17:54
I think you're somewhat mistaken, Ross, or at least that your comment is possibly misleading; I'll add something to my answer to clarify.
– John Hughes
Jul 19 at 18:41
I think you're somewhat mistaken, Ross, or at least that your comment is possibly misleading; I'll add something to my answer to clarify.
– John Hughes
Jul 19 at 18:41
add a comment |Â
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