dH/dt

[c4l3b]

Hi, currently studying some Math over the summer. I came across this question;

I need to find $\displaystyle dH/dt$

\(\displaystyle H = sin(xy) - 3y\^{}^2\)

Where $\displaystyle x = 3t + 1$ ; y = e^-t

I would really appreciate, if someone show me how to calculate the formula.

 

[Deleted member 4993]

Hi, currently studying some Math over the summer. I came across this question;

I need to find $\displaystyle dH/dt$

\(\displaystyle H = sin(xy) - 3y\^{}^2\)

Where $\displaystyle x = 3t + 1$ ; y = e^-t

I would really appreciate, if someone show me how to calculate the formula.

I'll do a similar but different problem for you:

$\displaystyle H = tan(xy) - 3x^3$

Where $\displaystyle x = ln(t)$ ; y = e^-t

We know:
$\displaystyle \frac{dx}{dt} = \frac{1}{t}$

$\displaystyle \frac{dy}{dt} = -e^{-t}$

$\displaystyle \frac{\partial H}{\partial x} \, = \, y \cdot sec^2(xy) \, - 9x^2$

$\displaystyle \frac{\partial H}{\partial y} \, = \, x \cdot sec^2(xy) \,$

$\displaystyle dH \, = \, \frac{\partial H}{\partial x} dx + \, \frac{\partial H}{\partial y} dy$

$\displaystyle \frac{dH}{dt} \, = \, \frac{\partial H}{\partial x} \frac{dx}{dt} + \, \frac{\partial H}{\partial y}\frac{dy}{dt}$

$\displaystyle \frac{dH}{dt} \, = \, [y \cdot sec^2(xy) \, - 9x^2]\cdot [\frac{1}{t}] \, + \, x \cdot sec^2(xy)\cdot [-e^{-t}]$

Follow the same steps....

 

[c4l3b]

Hi, currently studying some Math over the summer. I came across this question;

I need to find $\displaystyle dH/dt$

\(\displaystyle H = sin(xy) - 3y\^{}^2\)

Where $\displaystyle x = 3t + 1$ ; y = e^-t

I would really appreciate, if someone show me how to calculate the formula.

I'll do a similar but different problem for you:

$\displaystyle H = tan(xy) - 3x^3$

Where $\displaystyle x = ln(t)$ ; y = e^-t

We know:
$\displaystyle \frac{dx}{dt} = \frac{1}{t}$

$\displaystyle \frac{dy}{dt} = -e^{-t}$

$\displaystyle \frac{\partial H}{\partial x} \, = \, y \cdot sec^2(xy) \, - 9x^2$

$\displaystyle \frac{\partial H}{\partial y} \, = \, x \cdot sec^2(xy) \,$

$\displaystyle dH \, = \, \frac{\partial H}{\partial x} dx + \, \frac{\partial H}{\partial y} dy$

$\displaystyle \frac{dH}{dt} \, = \, \frac{\partial H}{\partial x} \frac{dx}{dt} + \, \frac{\partial H}{\partial y}\frac{dy}{dt}$

$\displaystyle \frac{dH}{dt} \, = \, [y \cdot sec^2(xy) \, - 9x^2]\cdot [\frac{1}{t}] \, + \, x \cdot sec^2(xy)\cdot [-e^{-t}]$

Follow the same steps....

Ok so; find $\displaystyle dH/dt$

$\displaystyle H = sin(xy) - 3y^2$

Where $\displaystyle x = 3t + 1$ ; y = e^-t


$\displaystyle \frac{dx}{dt} = 3$

$\displaystyle \frac{dy}{dt} = -e^{-t}$

$\displaystyle \frac{\partial H}{\partial x} \, = \, y \ cos(xy) \, - 6y$

$\displaystyle \frac{\partial H}{\partial y} \, = \, x \ cos(xy) \,$

$\displaystyle \frac{dH}{dt} \, = \, \frac{\partial H}{\partial x} \frac{dx}{dt} + \, \frac{\partial H}{\partial y}\frac{dy}{dt}$

$\displaystyle \frac{dH}{dt} \, = \, [y \cos(xy) \, - 6y]\cdot[3] \, + \, x \cos(xy)\cdot [-e^{-t}]$

Does anybody concur with my answer?

 

[Deleted member 4993]

Ok so; find $\displaystyle dH/dt$

$\displaystyle H = sin(xy) - 3y^2$

Where $\displaystyle x = 3t + 1$ ; y = e^-t


$\displaystyle \frac{dx}{dt} = 3$

$\displaystyle \frac{dy}{dt} = -e^{-t}$

$\displaystyle \frac{\partial H}{\partial x} \, = \, y \ cos(xy) \, - 6y$

$\displaystyle \frac{\partial H}{\partial y} \, = \, x \ cos(xy) \,$

$\displaystyle \frac{dH}{dt} \, = \, \frac{\partial H}{\partial x} \frac{dx}{dt} + \, \frac{\partial H}{\partial y}\frac{dy}{dt}$

$\displaystyle \frac{dH}{dt} \, = \, [y \cos(xy) \, - 6y]\cdot[3] \, - \, x \cos(xy)\cdot [e^{-t}]$

Does anybody concur with my answer?<<< By jove... I believe you conquered it!!!

 

[c4l3b]

Ok so; find $\displaystyle dH/dt$

$\displaystyle H = sin(xy) - 3y^2$

Where $\displaystyle x = 3t + 1$ ; y = e^-t


$\displaystyle \frac{dx}{dt} = 3$

$\displaystyle \frac{dy}{dt} = -e^{-t}$

$\displaystyle \frac{\partial H}{\partial x} \, = \, y \ cos(xy) \, - 6y$

$\displaystyle \frac{\partial H}{\partial y} \, = \, x \ cos(xy) \,$

$\displaystyle \frac{dH}{dt} \, = \, \frac{\partial H}{\partial x} \frac{dx}{dt} + \, \frac{\partial H}{\partial y}\frac{dy}{dt}$

$\displaystyle \frac{dH}{dt} \, = \, [y \cos(xy) \, - 6y]\cdot[3] \, - \, x \cos(xy)\cdot [e^{-t}]$

Does anybody concur with my answer?<<< By jove... I believe you conquered it!!!

Since this is partial differentiation solved by the chain rule method, which clues do I need to find to help me determine that I need to use the chain rule-

- would sin(xy) be a clue?
- note that x and y are functions of t?

Btw you can also use vector-matrix form, please correct me if I am wrong.

 

[Deleted member 4993]

[quote="c4l3b]Since this is partial differentiation solved by the chain rule method, which clues do I need to find to help me determine that I need to use the chain rule-

You can call it using chain rule but what you are really using is: when "C" is a constant

$\displaystyle \frac{d(Cu)}{dv} \, = \, C\cdot\frac{du}{dv}$

Ofcourse you can derive the above from "chain-rule" or from fundamental definition of differentiation"

- would sin(xy) be a clue?
- note that x and y are functions of t?

Btw you can also use vector-matrix form, <<< Why would you want to do that here??

please correct me if I am wrong.[/quote]

 

[c4l3b]

[quote="c4l3b]Since this is partial differentiation solved by the chain rule method, which clues do I need to find to help me determine that I need to use the chain rule-

You can call it using chain rule but what you are really using is: when "C" is a constant

$\displaystyle \frac{d(Cu)}{dv} \, = \, C\cdot\frac{du}{dv}$

Ofcourse you can derive the above from "chain-rule" or from fundamental definition of differentiation"

- would sin(xy) be a clue?
- note that x and y are functions of t?

Btw you can also use vector-matrix form, <<< Why would you want to do that here??

please correct me if I am wrong.

[/quote][/quote][/quote]


vector-matrix form.... just saying :lol:

Thanks for your help Subhotosh