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2 years ago

13

A satellite dish shaped like a paraboloid, has diameter 2.4 ft and depth 0.9 ft. If the receiver is placed at the focus, how far

should the receiver be from the vertex? Send help, we need to pass it before 12 midnight today

1 answer:

GREYUIT [131]2 years ago

7 0

Answer:

0.4

Step-by-step explanation:

x= $\frac{d}{2} =\frac{2.4}{2} =1.2\\$

$p=\frac{x^{2} }{4y}$

$p=\frac{(1.2)^{2} }{4(0.9)}$

p=0.4

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The temperature at a point (x, y) on a flat metal plate is given by T(x, y) = 88/(2 + x2 + y2), where T is measured in °C and x,

-Dominant- [34]

Answer:

$D_uT(3,1)=-\frac{44}{9}*\frac{1}{\sqrt{2} } \approx-3.46$

Step-by-step explanation:

To find the rate of change of temperature with respect to distance at the point (3, 1) in the x-direction and the y-direction we need to find the Directional Derivative of T(x,y). The definition of the directional derivative is given by:

$D_uT(x,y)=T_x(x,y)i+T_y(x,y)j$

Where i and j are the rectangular components of a unit vector. In this case, the problem don't give us additional information, so let's asume:

$i=\frac{1}{\sqrt{2} }$

$j=\frac{1}{\sqrt{2} }$

So, we need to find the partial derivative with respect to x and y:

In order to do the things easier let's make the next substitution:

$u=2+x^2+y^2$

and express T(x,y) as:

$T(x,y)=88*u^{-1}$

The partial derivative with respect to x is:

Using the chain rule:

$\frac{\partial u}{\partial x}=2x$

Hence:

$T_x(x,y)=88*(u^{-2})*\frac{\partial u}{\partial x}$

Symplying the expression and replacing the value of u:

$T_x(x,y)=\frac{-176x}{(2+x^2+y^2)^2}$

The partial derivative with respect to y is:

Using the chain rule:

$\frac{\partial u}{\partial y}=2y$

Hence:

$T_y(x,y)=88*(u^{-2})*\frac{\partial u}{\partial y}$

Symplying the expression and replacing the value of u:

$T_y(x,y)=\frac{-176y}{(2+x^2+y^2)^2}$

Therefore:

$D_uT(x,y)=(\frac{1}{\sqrt{2} } )*(\frac{-176x}{(2+x^2+y^2)^2} -\frac{176y}{(2+x^2+y^2)^2})$

Evaluating the point (3,1)

$D_uT(3,1)=(\frac{1}{\sqrt{2} } )*(\frac{-176(3)-176(1)}{(2+3^2+1^2)^2})=(\frac{1}{\sqrt{2} })* (-\frac{704}{144})=(\frac{1}{\sqrt{2} }) ( - \frac{44}{9})\approx -3.46$

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