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2 years ago

The table shows how an elevator 500 feet above the ground is descending at a steady rate.

2 answers:

8 0

<span>(Attached Image: https://media.education2020.com/evresources/3109/3109-03/3109-03-06/3109-03-06-assessment/3109-03-06...) The correct answer is: h(t) = -5t + 500; from the table it is visible that the time difference is 5 seconds; the sign before 500 must be be "+", for we're talking about 500 feet above the ground. </span>

Snezhnost [94]2 years ago

6 0

We are given that the elevator is descending, so the coefficient of t must be negative. The elevator's initial position is 500 feet above the ground, so this is a positive value of +500. Therefore the only choice that fits these is h(t) = -5t + 500.

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$\mathbf J=\begin{bmatrix}\dfrac{\partial u}{\partial x}&\dfrac{\partial v}{\partial x}\\\\\dfrac{\partial u}{\partial y}&\dfrac{\partial v}{\partial y}\end{bmatrix}=\begin{bmatrix}y&x\\\\\dfrac1y&-\dfrac x{y^2}\end{bmatrix}$

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Note that we need to use the Jacobian in the other direction; that is, we've computed

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we need to take the reciprocal of the Jacobian above.

The integral then changes to

$\displaystyle\iint_{\mathcal W_{(x,y)}}e^{xy}\,\mathrm dx\,\mathrm dy=\iint_{\mathcal W_{(u,v)}}\dfrac{e^u}{|\det\mathbf J|}\,\mathrm du\,\mathrm dv$
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3 years ago

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3 years ago

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