Point a is at (6,-6) and point c is at (-6, -2)

Let w(s,t)=f(u(s,t),v(s,t)) where u(1,0)=−6,∂u∂s(1,0)=5,∂u∂1(1,0)=7 v(1,0)=−8,∂v∂s(1,0)=−8,∂v∂t(1,0)=6 ∂f∂u(−6,−8)=−1,∂f∂v(−6,−8

Blababa [14]

$w(s,t)=f(u(s,t),v(s,t))$

From the given set of conditions, it's likely that you are asked to find the values of $\dfrac{\partial w}{\partial s}$ and $\dfrac{\partial w}{\partial t}$ at the point $(s,t)=(1,0)$ .

By the chain rule, the partial derivative with respect to $s$ is

$\dfrac{\partial w}{\partial s}=\dfrac{\partial f}{\partial u}\dfrac{\partial u}{\partial s}+\dfrac{\partial f}{\partial v}\dfrac{\partial v}{\partial s}$

and so at the point $(1,0)$ , we have

$\dfrac{\partial w}{\partial s}\bigg|_{(s,t)=(1,0)}=\dfrac{\partial f}{\partial 
u}\bigg|_{(u,v)=(-6,-8)}\dfrac{\partial u}{\partial s}\bigg|_{(s,t)=(1,0)}+\dfrac{\partial f}{\partial 
v}\bigg|_{(u,v)=(-6,-8)}\dfrac{\partial v}{\partial s}\bigg|_{(s,t)=(1,0)}$
$\dfrac{\partial w}{\partial s}\bigg|_{(s,t)=(1,0)}=(-1)(5)+(2)(-8)=-21$

Similarly, the partial derivative with respect to $t$ would be found via

$\dfrac{\partial w}{\partial t}\bigg|_{(s,t)=(1,0)}=\dfrac{\partial f}{\partial 
u}\bigg|_{(u,v)=(-6,-8)}\dfrac{\partial u}{\partial t}\bigg|_{(s,t)=(1,0)}+\dfrac{\partial f}{\partial 
v}\bigg|_{(u,v)=(-6,-8)}\dfrac{\partial v}{\partial t}\bigg|_{(s,t)=(1,0)}$
$\dfrac{\partial w}{\partial t}\bigg|_{(s,t)=(1,0)}=(-1)(7)+(2)(6)=5$

6 0

3 years ago

A baker used 3/16 cup of sugar. Then she used another 1/2 cup of sugar. How much sugar did she use in all?

Fittoniya [83]

Answer:

11/16 :)

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

1. The graph below shows the past 9 months of sales for the Carver Cars company. у Carver Cars Sales Car Sales (in dollars) 200,

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