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Perpendicular tangents of parabola and possible functions g(x)
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I am taking a calculus class and I've already asked for help but I didn't really get it :(
Can someone help me start this? Thanks!
Let $f(x)$ and $g(x)$ be functions with domain $(0,infty).$ Suppose $f(x)=x^2$ and the tangent line to $f(x)$ at $x=a$ is perpendicular to the tangent line to $g(x)$ at $x=a$ for all positive real numbers $a$. Find all possible functions $g(x).$
Here's what I don't get/don't know what supposed to do: I tried graphing this on desmos and found it REALLY hard to find functions that would work and I'm guessing that the tangents would have to be close to (0,0). I'm assuming that g(x) is dependent on f(x). So, do I need to find functions for f(x) or is there something else?
calculus real-analysis derivatives
asked Jul 18 at 23:57
ninjagirl
1427
add a comment |Â
up vote
1
down vote
favorite
I am taking a calculus class and I've already asked for help but I didn't really get it :(
Can someone help me start this? Thanks!
Let $f(x)$ and $g(x)$ be functions with domain $(0,infty).$ Suppose $f(x)=x^2$ and the tangent line to $f(x)$ at $x=a$ is perpendicular to the tangent line to $g(x)$ at $x=a$ for all positive real numbers $a$. Find all possible functions $g(x).$
Here's what I don't get/don't know what supposed to do: I tried graphing this on desmos and found it REALLY hard to find functions that would work and I'm guessing that the tangents would have to be close to (0,0). I'm assuming that g(x) is dependent on f(x). So, do I need to find functions for f(x) or is there something else?
calculus real-analysis derivatives
asked Jul 18 at 23:57
ninjagirl
1427
Take the problem in steps. (1) What is the slope of the line tangent to $f(x)$ at $x=a$? (2) What's the slope of any line perpendicular to that line? (3) What does $g(x)$ have to be so that the slope of its tangent line at $x=a$ is the result of (2)?
– Blue
Jul 19 at 0:04
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am taking a calculus class and I've already asked for help but I didn't really get it :(
Can someone help me start this? Thanks!
Let $f(x)$ and $g(x)$ be functions with domain $(0,infty).$ Suppose $f(x)=x^2$ and the tangent line to $f(x)$ at $x=a$ is perpendicular to the tangent line to $g(x)$ at $x=a$ for all positive real numbers $a$. Find all possible functions $g(x).$
Here's what I don't get/don't know what supposed to do: I tried graphing this on desmos and found it REALLY hard to find functions that would work and I'm guessing that the tangents would have to be close to (0,0). I'm assuming that g(x) is dependent on f(x). So, do I need to find functions for f(x) or is there something else?
calculus real-analysis derivatives
asked Jul 18 at 23:57
ninjagirl
1427
I am taking a calculus class and I've already asked for help but I didn't really get it :(
Can someone help me start this? Thanks!
Let $f(x)$ and $g(x)$ be functions with domain $(0,infty).$ Suppose $f(x)=x^2$ and the tangent line to $f(x)$ at $x=a$ is perpendicular to the tangent line to $g(x)$ at $x=a$ for all positive real numbers $a$. Find all possible functions $g(x).$
Here's what I don't get/don't know what supposed to do: I tried graphing this on desmos and found it REALLY hard to find functions that would work and I'm guessing that the tangents would have to be close to (0,0). I'm assuming that g(x) is dependent on f(x). So, do I need to find functions for f(x) or is there something else?
calculus real-analysis derivatives
asked Jul 18 at 23:57
ninjagirl
1427
asked Jul 18 at 23:57
ninjagirl
1427
asked Jul 18 at 23:57
ninjagirl
1427
asked Jul 18 at 23:57
ninjagirl
1427
1427
Take the problem in steps. (1) What is the slope of the line tangent to $f(x)$ at $x=a$? (2) What's the slope of any line perpendicular to that line? (3) What does $g(x)$ have to be so that the slope of its tangent line at $x=a$ is the result of (2)?
– Blue
Jul 19 at 0:04
add a comment |Â
Take the problem in steps. (1) What is the slope of the line tangent to $f(x)$ at $x=a$? (2) What's the slope of any line perpendicular to that line? (3) What does $g(x)$ have to be so that the slope of its tangent line at $x=a$ is the result of (2)?
– Blue
Jul 19 at 0:04
Take the problem in steps. (1) What is the slope of the line tangent to $f(x)$ at $x=a$? (2) What's the slope of any line perpendicular to that line? (3) What does $g(x)$ have to be so that the slope of its tangent line at $x=a$ is the result of (2)?
– Blue
Jul 19 at 0:04
Take the problem in steps. (1) What is the slope of the line tangent to $f(x)$ at $x=a$? (2) What's the slope of any line perpendicular to that line? (3) What does $g(x)$ have to be so that the slope of its tangent line at $x=a$ is the result of (2)?
– Blue
Jul 19 at 0:04
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
First, let's find the slope of the tangent line to $f(x)$ at $x = a$. The derivative of $x^2$ is $2x$, so the slope of the tangent line is $2a$.
Because the two tangent lines are perpendicular, the slope of the tangent line to $g(x)$ is $-frac12a$. Basically, we want to find all the functions $g(x)$ can be for its derivative at $a$ to be $-frac12a$. So just take the antiderivative of $-frac12a$. We can replace $a$ with $x$ to get $-frac12(x)^-1$, and the antiderivative is $$-fraclnx2$$ Adding $C$ as a constant (moving a function up and down will not change its derivative), we find that $$-fraclnx2+C$$ is the correct answer.
answered Jul 19 at 0:54
RayDansh
882214
Note that we're looking for a single function $g(x)$ with the perpendicular-tangent property "for all positive real numbers $a$". That is, we want a function $g(x)$ such that $g^prime(a) = -1/(2a)$ for all $a$. (In other symbols, $g^prime(x) = -1/(2x)$.) In effect, $a$ does change.
– Blue
Jul 19 at 1:01
@Blue I see your point. Let me investigate.
– RayDansh
Jul 19 at 1:02
@Blue Would you mind checking my solution again?
– RayDansh
Jul 19 at 1:05
It's correct now ... but you don't need the absolute value, since the domain is assumed to be the positive reals.
– Blue
Jul 19 at 1:06
1
@Blue Thanks. It should be updated now.
– RayDansh
Jul 19 at 1:07
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
First, let's find the slope of the tangent line to $f(x)$ at $x = a$. The derivative of $x^2$ is $2x$, so the slope of the tangent line is $2a$.
Because the two tangent lines are perpendicular, the slope of the tangent line to $g(x)$ is $-frac12a$. Basically, we want to find all the functions $g(x)$ can be for its derivative at $a$ to be $-frac12a$. So just take the antiderivative of $-frac12a$. We can replace $a$ with $x$ to get $-frac12(x)^-1$, and the antiderivative is $$-fraclnx2$$ Adding $C$ as a constant (moving a function up and down will not change its derivative), we find that $$-fraclnx2+C$$ is the correct answer.
answered Jul 19 at 0:54
RayDansh
882214
Note that we're looking for a single function $g(x)$ with the perpendicular-tangent property "for all positive real numbers $a$". That is, we want a function $g(x)$ such that $g^prime(a) = -1/(2a)$ for all $a$. (In other symbols, $g^prime(x) = -1/(2x)$.) In effect, $a$ does change.
– Blue
Jul 19 at 1:01
@Blue I see your point. Let me investigate.
– RayDansh
Jul 19 at 1:02
@Blue Would you mind checking my solution again?
– RayDansh
Jul 19 at 1:05
It's correct now ... but you don't need the absolute value, since the domain is assumed to be the positive reals.
– Blue
Jul 19 at 1:06
1
@Blue Thanks. It should be updated now.
– RayDansh
Jul 19 at 1:07
add a comment |Â
up vote
2
down vote
accepted
First, let's find the slope of the tangent line to $f(x)$ at $x = a$. The derivative of $x^2$ is $2x$, so the slope of the tangent line is $2a$.
Because the two tangent lines are perpendicular, the slope of the tangent line to $g(x)$ is $-frac12a$. Basically, we want to find all the functions $g(x)$ can be for its derivative at $a$ to be $-frac12a$. So just take the antiderivative of $-frac12a$. We can replace $a$ with $x$ to get $-frac12(x)^-1$, and the antiderivative is $$-fraclnx2$$ Adding $C$ as a constant (moving a function up and down will not change its derivative), we find that $$-fraclnx2+C$$ is the correct answer.
answered Jul 19 at 0:54
RayDansh
882214
Note that we're looking for a single function $g(x)$ with the perpendicular-tangent property "for all positive real numbers $a$". That is, we want a function $g(x)$ such that $g^prime(a) = -1/(2a)$ for all $a$. (In other symbols, $g^prime(x) = -1/(2x)$.) In effect, $a$ does change.
– Blue
Jul 19 at 1:01
@Blue I see your point. Let me investigate.
– RayDansh
Jul 19 at 1:02
@Blue Would you mind checking my solution again?
– RayDansh
Jul 19 at 1:05
It's correct now ... but you don't need the absolute value, since the domain is assumed to be the positive reals.
– Blue
Jul 19 at 1:06
1
@Blue Thanks. It should be updated now.
– RayDansh
Jul 19 at 1:07
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
First, let's find the slope of the tangent line to $f(x)$ at $x = a$. The derivative of $x^2$ is $2x$, so the slope of the tangent line is $2a$.
Because the two tangent lines are perpendicular, the slope of the tangent line to $g(x)$ is $-frac12a$. Basically, we want to find all the functions $g(x)$ can be for its derivative at $a$ to be $-frac12a$. So just take the antiderivative of $-frac12a$. We can replace $a$ with $x$ to get $-frac12(x)^-1$, and the antiderivative is $$-fraclnx2$$ Adding $C$ as a constant (moving a function up and down will not change its derivative), we find that $$-fraclnx2+C$$ is the correct answer.
answered Jul 19 at 0:54
RayDansh
882214
First, let's find the slope of the tangent line to $f(x)$ at $x = a$. The derivative of $x^2$ is $2x$, so the slope of the tangent line is $2a$.
Because the two tangent lines are perpendicular, the slope of the tangent line to $g(x)$ is $-frac12a$. Basically, we want to find all the functions $g(x)$ can be for its derivative at $a$ to be $-frac12a$. So just take the antiderivative of $-frac12a$. We can replace $a$ with $x$ to get $-frac12(x)^-1$, and the antiderivative is $$-fraclnx2$$ Adding $C$ as a constant (moving a function up and down will not change its derivative), we find that $$-fraclnx2+C$$ is the correct answer.
answered Jul 19 at 0:54
RayDansh
882214
edited Jul 19 at 1:07
answered Jul 19 at 0:54
RayDansh
882214
answered Jul 19 at 0:54
RayDansh
882214
answered Jul 19 at 0:54
RayDansh
882214
882214
Note that we're looking for a single function $g(x)$ with the perpendicular-tangent property "for all positive real numbers $a$". That is, we want a function $g(x)$ such that $g^prime(a) = -1/(2a)$ for all $a$. (In other symbols, $g^prime(x) = -1/(2x)$.) In effect, $a$ does change.
– Blue
Jul 19 at 1:01
@Blue I see your point. Let me investigate.
– RayDansh
Jul 19 at 1:02
@Blue Would you mind checking my solution again?
– RayDansh
Jul 19 at 1:05
It's correct now ... but you don't need the absolute value, since the domain is assumed to be the positive reals.
– Blue
Jul 19 at 1:06
1
@Blue Thanks. It should be updated now.
– RayDansh
Jul 19 at 1:07
add a comment |Â
Note that we're looking for a single function $g(x)$ with the perpendicular-tangent property "for all positive real numbers $a$". That is, we want a function $g(x)$ such that $g^prime(a) = -1/(2a)$ for all $a$. (In other symbols, $g^prime(x) = -1/(2x)$.) In effect, $a$ does change.
– Blue
Jul 19 at 1:01
@Blue I see your point. Let me investigate.
– RayDansh
Jul 19 at 1:02
@Blue Would you mind checking my solution again?
– RayDansh
Jul 19 at 1:05
It's correct now ... but you don't need the absolute value, since the domain is assumed to be the positive reals.
– Blue
Jul 19 at 1:06
1
@Blue Thanks. It should be updated now.
– RayDansh
Jul 19 at 1:07
Note that we're looking for a single function $g(x)$ with the perpendicular-tangent property "for all positive real numbers $a$". That is, we want a function $g(x)$ such that $g^prime(a) = -1/(2a)$ for all $a$. (In other symbols, $g^prime(x) = -1/(2x)$.) In effect, $a$ does change.
– Blue
Jul 19 at 1:01
Note that we're looking for a single function $g(x)$ with the perpendicular-tangent property "for all positive real numbers $a$". That is, we want a function $g(x)$ such that $g^prime(a) = -1/(2a)$ for all $a$. (In other symbols, $g^prime(x) = -1/(2x)$.) In effect, $a$ does change.
– Blue
Jul 19 at 1:01
@Blue I see your point. Let me investigate.
– RayDansh
Jul 19 at 1:02
@Blue I see your point. Let me investigate.
– RayDansh
Jul 19 at 1:02
@Blue Would you mind checking my solution again?
– RayDansh
Jul 19 at 1:05
@Blue Would you mind checking my solution again?
– RayDansh
Jul 19 at 1:05
It's correct now ... but you don't need the absolute value, since the domain is assumed to be the positive reals.
– Blue
Jul 19 at 1:06
It's correct now ... but you don't need the absolute value, since the domain is assumed to be the positive reals.
– Blue
Jul 19 at 1:06
1
1
@Blue Thanks. It should be updated now.
– RayDansh
Jul 19 at 1:07
@Blue Thanks. It should be updated now.
– RayDansh
Jul 19 at 1:07
add a comment |Â
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Take the problem in steps. (1) What is the slope of the line tangent to $f(x)$ at $x=a$? (2) What's the slope of any line perpendicular to that line? (3) What does $g(x)$ have to be so that the slope of its tangent line at $x=a$ is the result of (2)?
– Blue
Jul 19 at 0:04