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Veseljchak [2.6K]
2 years ago
10

Find the midpoint of A and B where A has coordinates (-3,-5) and B has coordinates (4,4).

Mathematics
1 answer:
snow_lady [41]2 years ago
8 0

Answer:

(\frac{1}{2}, \frac{-1}{2})

Step-by-step explanation:

Use the midpoint formula to solve.

1) Given 2 points: (x_{1}, y_1) and (x_2, y_2)

The midpoint formula would be (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}).

2) Based on this, we can easily find the midpoint of AB.

  • -3 would be x_1\\
  • -5 would be y_1
  • 4 would be x_2  
  • the next 4 would be y_2

3) With this information, we can solve for the midpoint.  

(\frac{-3+4}{2}, \frac{-5+4}{2}) --> (\frac{1}{2}, \frac{-1}{2}) --> This is your answer.

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Use the Chain Rule to find the indicated partial derivatives. z = x^4 + xy^3, x = uv^4 + w^3, y = u + ve^w Find : ∂z/∂u , ∂z/∂v
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I'll use subscript notation for brevity, i.e. \frac{\partial f}{\partial x}=f_x.

By the chain rule,

z_u=z_xx_u+z_yy_u

z_v=z_xx_v+z_yy_v

z_w=z_xx_w+z_yy_w

We have

z=x^4+xy^3\implies\begin{cases}z_x=4x^3+y^3\\z_y=3xy^2\end{cases}

and

\begin{cases}x=uv^4+w^3\\y=u+ve^w\end{cases}\implies\begin{cases}x_u=v^4\\x_v=4uv^3\\x_w=3w^2\\y_u=1\\y_v=e^w\\y_w=ve^w\end{cases}

When u=1,v=1,w=0, we have

\begin{cases}x(1,1,0)=1\\y(1,1,0)=2\end{cases}\implies\begin{cases}z_x(1,2)=12\\z_y(1,2)=12\end{cases}

and the partial derivatives take on values of

\begin{cases}x_u(1,1,0)=1\\x_v(1,1,0)=4\\x_w(1,1,0)=0\\y_u(1,1,0)=1\\y_v(1,1,0)=1\\y_w(1,1,0)=1\end{cases}

So we end up with

\boxed{\begin{cases}z_u(1,1,0)=24\\z_v(1,1,0)=60\\z_w(1,1,0)=12\end{cases}}

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