Chain Rule To Find Dz/dt

Question

Note the multivariable chain rule: $$\frac{dz}{dt}=\frac{\partial z}{\partial x}\cdot \frac{dx}{dt}+\frac{\fractional z}{\partial y}\cdot \frac{dy}{dt} \tag{one}$$ Simply evaluate the ordinary and partial derivatives, and substitute them into the equation above. You should have a consequence for $\frac{dz}{dt}$ in terms of $x,y$ and $t$. And so, substitute the parametrizations for $x(t)$ and $y(t)$ to obtain an expression for $\frac{dz}{dt}$ strictly in terms of $t$ equally done on the outset example of the link I provided.

Let'south get-go by evaluating $\frac{\partial z}{\partial x}$. Since nosotros must care for the variable $y$ equally a constant as a effect of partial differentiation, $e^{2y}$ is as well treated as one. Therefore: $$\frac{\partial z}{\fractional x}=3x^ii\cdot e^{2y}$$ Tin can you continue?

Accepted Answer · 2017-03-19 21:08:02Z

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Note the multivariable chain rule: $$\frac{dz}{dt}=\frac{\partial z}{\partial x}\cdot \frac{dx}{dt}+\frac{\fractional z}{\partial y}\cdot \frac{dy}{dt} \tag{one}$$ Simply evaluate the ordinary and partial derivatives, and substitute them into the equation above. You should have a consequence for $\frac{dz}{dt}$ in terms of $x,y$ and $t$. And so, substitute the parametrizations for $x(t)$ and $y(t)$ to obtain an expression for $\frac{dz}{dt}$ strictly in terms of $t$ equally done on the outset example of the link I provided.

Let'south get-go by evaluating $\frac{\partial z}{\partial x}$. Since nosotros must care for the variable $y$ equally a constant as a effect of partial differentiation, $e^{2y}$ is as well treated as one. Therefore: $$\frac{\partial z}{\fractional x}=3x^ii\cdot e^{2y}$$ Tin can you continue?

answered Mar nineteen, 2017 at 21:08

projectilemotionprojectilemotion

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$\begingroup$ Thank you I'll try this and get back to you $\endgroup$

Mar nineteen, 2017 at 21:31
$\begingroup$ Okay I calculated the partial and normal derivites but I dont understand where to plug them in. The derivitive dx/dt =2, dy/dt= 2t, partial x = 3x^2x east^(2y) and partial y as 2e^(2y) $\endgroup$

Mar xix, 2017 at 23:23
$\begingroup$ @Clueless Plug them in to equation $(1)$. Out of curiosity, what did you evaluate each of the derivatives to be? $\endgroup$

Mar 19, 2017 at 23:23
$\begingroup$ The derivitive dx/dt =2, dy/dt= 2t, partial 10 = 3x^2x e^(2y) and partial y=2e^(2y) $\endgroup$

Mar 19, 2017 at 23:27
$\begingroup$ @Clueless Did y'all make a typo? For $z=x^three\cdot e^{2y}$ you should obtain: $$\frac{\fractional z}{\partial ten}=3x^ii\cdot e^{2y}$$ $$\frac{\partial z}{\partial y}=2x^3\cdot e^{2y}$$ Otherwise, the balance are correct. $\endgroup$

Mar 19, 2017 at 23:30

Chain Rule To Find Dz/dt

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