• anonymous
In Problem 7 in the problem set 1part1 can you explain how delta x × k1 × t = k2 × delta t from an intuitive sense?
MIT 18.03SC Differential Equations
  • jamiebookeater
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  • anonymous
I don't understand this problem very well either, so maybe someone others can reply too, but here is what I am thinking. As the solutions explain, k1(t) is the height of the snow at time elapsed = t. So k1(t) is the total height of the snow accumulated up to time t. In a certain period of time (delta t), the plow will move by amount delta(x). Multiplying the height of the snow and the horizontal movement of the plow gives us the vertical cross sectional area plowed. (You can imagine a rectangle with the vertical height of the snow (k1) and the horizontal plow distance (deltax).) If we multiply this by L, the length of the plow, we get the rectangular prism containing the volume of snow plowed during this time. That is the right hand side of the equation (the solution does not include L, but as L is constant, it may be included in the constant k1). On the left hand side of the equation we have k2delta(t). This will also give the total volume of snow plowed during the time interval delta t. The rate of plowing in m^3/hr is given by k2. So K2delta (t) = (m^3/hour) (hour) = m^3. So k2delta(t) gives us the total meters cubed (i.e. the total volume) of snow plowed during the time delta(t). So, 2 ways to get volume: distance and height and length, or rate and time.
  • anonymous
I'm not sure where you're finding this question. Can you either direct me to where you have the question, or reproduce it completely?

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