Is this the correct method to compute the directional derivative of $X_p[f]$?












0














Say we have a smooth function



$f(x,y,z)=z^2+7x^2e^{yz}$



and a vector field



$X(x,y,z)=(-x,y,z)$



Now say we want to calculate the directional derivative $X(p)[f]$ at the point $p=(0,1,2)$. Then is this the correct method :



$X(p)[f]=tfrac{d}{dt}|_{t=0}f(alpha(t))$



$alpha(t)=p+tX=(0,1,2)+t(-x,y,z)=(0,1,2)+t(0,1,2)=(0,1+t,2+2t)$



$therefore f(alpha(t))=(2+2t)^2=4+8t+4t^2$



$tfrac{df(alpha(t)}{dt}=8+8t$



so



$tfrac{df(alpha(t)}{dt}|_{t=0}=8$



which is just a scalar number.



Note: the reason I'm asking is because i'm studying for an exam an I wasn't sure whether $p+tx=(0,1,2)+t(-x,y,z)=(-tx,1+ty,2+2z)$ or whether it was what I stated above , above produces a scalar this method produces a smooth function.






differential-geometry







share|cite|improve this question
























  • You need any curve $alpha(t)$ with $alpha(0)=p$ and $alpha'(0)=X(p)$. In your second try, I don't know what $x$, $y$, and $z$ are. Do you?
    – Ted Shifrin
    Jan 4 at 19:19












  • @TedShifrin In the second try I just meant x,y,z to represent the vector field , but I left it arbitrary in that case , In the first try I calculated X at the point p and then x,y,z became (0,1,2). So the first try is right then ?
    – can'tcauchy
    Jan 4 at 20:43










  • Well, you could use the flow of the vector field $X$, in which case $alpha(t)$ will not be a line but instead a curve with $alpha'(t) = X(alpha(t))$ instead.
    – Ted Shifrin
    Jan 4 at 20:46










  • @TedShifrin I'm sorry but I don't really follow you, Ive only had one course on introductory differential geometry , this method was just the one we saw in class so I think its safer to just stick with it as long as its correct
    – can'tcauchy
    Jan 4 at 20:55










  • The definition of directional derivative you get in multivariable calculus is what you wrote down — you do the instantaneous rate of change (at the point) along a line through the point in the given direction.
    – Ted Shifrin
    Jan 4 at 21:08
















0














Say we have a smooth function



$f(x,y,z)=z^2+7x^2e^{yz}$



and a vector field



$X(x,y,z)=(-x,y,z)$



Now say we want to calculate the directional derivative $X(p)[f]$ at the point $p=(0,1,2)$. Then is this the correct method :



$X(p)[f]=tfrac{d}{dt}|_{t=0}f(alpha(t))$



$alpha(t)=p+tX=(0,1,2)+t(-x,y,z)=(0,1,2)+t(0,1,2)=(0,1+t,2+2t)$



$therefore f(alpha(t))=(2+2t)^2=4+8t+4t^2$



$tfrac{df(alpha(t)}{dt}=8+8t$



so



$tfrac{df(alpha(t)}{dt}|_{t=0}=8$



which is just a scalar number.



Note: the reason I'm asking is because i'm studying for an exam an I wasn't sure whether $p+tx=(0,1,2)+t(-x,y,z)=(-tx,1+ty,2+2z)$ or whether it was what I stated above , above produces a scalar this method produces a smooth function.






differential-geometry







share|cite|improve this question
























  • You need any curve $alpha(t)$ with $alpha(0)=p$ and $alpha'(0)=X(p)$. In your second try, I don't know what $x$, $y$, and $z$ are. Do you?
    – Ted Shifrin
    Jan 4 at 19:19












  • @TedShifrin In the second try I just meant x,y,z to represent the vector field , but I left it arbitrary in that case , In the first try I calculated X at the point p and then x,y,z became (0,1,2). So the first try is right then ?
    – can'tcauchy
    Jan 4 at 20:43










  • Well, you could use the flow of the vector field $X$, in which case $alpha(t)$ will not be a line but instead a curve with $alpha'(t) = X(alpha(t))$ instead.
    – Ted Shifrin
    Jan 4 at 20:46










  • @TedShifrin I'm sorry but I don't really follow you, Ive only had one course on introductory differential geometry , this method was just the one we saw in class so I think its safer to just stick with it as long as its correct
    – can'tcauchy
    Jan 4 at 20:55










  • The definition of directional derivative you get in multivariable calculus is what you wrote down — you do the instantaneous rate of change (at the point) along a line through the point in the given direction.
    – Ted Shifrin
    Jan 4 at 21:08














0












0








0







Say we have a smooth function



$f(x,y,z)=z^2+7x^2e^{yz}$



and a vector field



$X(x,y,z)=(-x,y,z)$



Now say we want to calculate the directional derivative $X(p)[f]$ at the point $p=(0,1,2)$. Then is this the correct method :



$X(p)[f]=tfrac{d}{dt}|_{t=0}f(alpha(t))$



$alpha(t)=p+tX=(0,1,2)+t(-x,y,z)=(0,1,2)+t(0,1,2)=(0,1+t,2+2t)$



$therefore f(alpha(t))=(2+2t)^2=4+8t+4t^2$



$tfrac{df(alpha(t)}{dt}=8+8t$



so



$tfrac{df(alpha(t)}{dt}|_{t=0}=8$



which is just a scalar number.



Note: the reason I'm asking is because i'm studying for an exam an I wasn't sure whether $p+tx=(0,1,2)+t(-x,y,z)=(-tx,1+ty,2+2z)$ or whether it was what I stated above , above produces a scalar this method produces a smooth function.






differential-geometry







share|cite|improve this question















Say we have a smooth function



$f(x,y,z)=z^2+7x^2e^{yz}$



and a vector field



$X(x,y,z)=(-x,y,z)$



Now say we want to calculate the directional derivative $X(p)[f]$ at the point $p=(0,1,2)$. Then is this the correct method :



$X(p)[f]=tfrac{d}{dt}|_{t=0}f(alpha(t))$



$alpha(t)=p+tX=(0,1,2)+t(-x,y,z)=(0,1,2)+t(0,1,2)=(0,1+t,2+2t)$



$therefore f(alpha(t))=(2+2t)^2=4+8t+4t^2$



$tfrac{df(alpha(t)}{dt}=8+8t$



so



$tfrac{df(alpha(t)}{dt}|_{t=0}=8$



which is just a scalar number.



Note: the reason I'm asking is because i'm studying for an exam an I wasn't sure whether $p+tx=(0,1,2)+t(-x,y,z)=(-tx,1+ty,2+2z)$ or whether it was what I stated above , above produces a scalar this method produces a smooth function.






differential-geometry




differential-geometry






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share|cite|improve this question













share|cite|improve this question




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edited Jan 4 at 19:14







can'tcauchy

















asked Jan 4 at 18:08









can'tcauchycan'tcauchy

975417




975417












  • You need any curve $alpha(t)$ with $alpha(0)=p$ and $alpha'(0)=X(p)$. In your second try, I don't know what $x$, $y$, and $z$ are. Do you?
    – Ted Shifrin
    Jan 4 at 19:19












  • @TedShifrin In the second try I just meant x,y,z to represent the vector field , but I left it arbitrary in that case , In the first try I calculated X at the point p and then x,y,z became (0,1,2). So the first try is right then ?
    – can'tcauchy
    Jan 4 at 20:43










  • Well, you could use the flow of the vector field $X$, in which case $alpha(t)$ will not be a line but instead a curve with $alpha'(t) = X(alpha(t))$ instead.
    – Ted Shifrin
    Jan 4 at 20:46










  • @TedShifrin I'm sorry but I don't really follow you, Ive only had one course on introductory differential geometry , this method was just the one we saw in class so I think its safer to just stick with it as long as its correct
    – can'tcauchy
    Jan 4 at 20:55










  • The definition of directional derivative you get in multivariable calculus is what you wrote down — you do the instantaneous rate of change (at the point) along a line through the point in the given direction.
    – Ted Shifrin
    Jan 4 at 21:08


















  • You need any curve $alpha(t)$ with $alpha(0)=p$ and $alpha'(0)=X(p)$. In your second try, I don't know what $x$, $y$, and $z$ are. Do you?
    – Ted Shifrin
    Jan 4 at 19:19












  • @TedShifrin In the second try I just meant x,y,z to represent the vector field , but I left it arbitrary in that case , In the first try I calculated X at the point p and then x,y,z became (0,1,2). So the first try is right then ?
    – can'tcauchy
    Jan 4 at 20:43










  • Well, you could use the flow of the vector field $X$, in which case $alpha(t)$ will not be a line but instead a curve with $alpha'(t) = X(alpha(t))$ instead.
    – Ted Shifrin
    Jan 4 at 20:46










  • @TedShifrin I'm sorry but I don't really follow you, Ive only had one course on introductory differential geometry , this method was just the one we saw in class so I think its safer to just stick with it as long as its correct
    – can'tcauchy
    Jan 4 at 20:55










  • The definition of directional derivative you get in multivariable calculus is what you wrote down — you do the instantaneous rate of change (at the point) along a line through the point in the given direction.
    – Ted Shifrin
    Jan 4 at 21:08
















You need any curve $alpha(t)$ with $alpha(0)=p$ and $alpha'(0)=X(p)$. In your second try, I don't know what $x$, $y$, and $z$ are. Do you?
– Ted Shifrin
Jan 4 at 19:19






You need any curve $alpha(t)$ with $alpha(0)=p$ and $alpha'(0)=X(p)$. In your second try, I don't know what $x$, $y$, and $z$ are. Do you?
– Ted Shifrin
Jan 4 at 19:19














@TedShifrin In the second try I just meant x,y,z to represent the vector field , but I left it arbitrary in that case , In the first try I calculated X at the point p and then x,y,z became (0,1,2). So the first try is right then ?
– can'tcauchy
Jan 4 at 20:43




@TedShifrin In the second try I just meant x,y,z to represent the vector field , but I left it arbitrary in that case , In the first try I calculated X at the point p and then x,y,z became (0,1,2). So the first try is right then ?
– can'tcauchy
Jan 4 at 20:43












Well, you could use the flow of the vector field $X$, in which case $alpha(t)$ will not be a line but instead a curve with $alpha'(t) = X(alpha(t))$ instead.
– Ted Shifrin
Jan 4 at 20:46




Well, you could use the flow of the vector field $X$, in which case $alpha(t)$ will not be a line but instead a curve with $alpha'(t) = X(alpha(t))$ instead.
– Ted Shifrin
Jan 4 at 20:46












@TedShifrin I'm sorry but I don't really follow you, Ive only had one course on introductory differential geometry , this method was just the one we saw in class so I think its safer to just stick with it as long as its correct
– can'tcauchy
Jan 4 at 20:55




@TedShifrin I'm sorry but I don't really follow you, Ive only had one course on introductory differential geometry , this method was just the one we saw in class so I think its safer to just stick with it as long as its correct
– can'tcauchy
Jan 4 at 20:55












The definition of directional derivative you get in multivariable calculus is what you wrote down — you do the instantaneous rate of change (at the point) along a line through the point in the given direction.
– Ted Shifrin
Jan 4 at 21:08




The definition of directional derivative you get in multivariable calculus is what you wrote down — you do the instantaneous rate of change (at the point) along a line through the point in the given direction.
– Ted Shifrin
Jan 4 at 21:08










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