course mth 173 4/8 3:00 003. `query 3
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Given Solution: ** You have to find the average rate of change between clock times t and t + `dt: ave rate of change = [ a (t+`dt)^2 + b (t+`dt) + c - ( a t^2 + b t + c ) ] / `dt = [ a t^2 + 2 a t `dt + a `dt^2 + b t + b `dt + c - ( a t^2 + b t + c ) ] / `dt = [ 2 a t `dt + a `dt^2 + b `dt ] / `dt = 2 a t + b t + a `dt. Now if `dt shrinks to a very small value the ave rate of change approaches y ' = 2 a t + b. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique rating #$&*ok ********************************************* Question: `q explain how we know that the depth function for rate-of-depth-change function y' = m t + b must be y = 1/2 m t^2 + b t + c, for some constant c, and explain the significance of the constant c. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since y' = mt + b, and we just established y' as equal to 2at + b, then 2at + b = mt + b. So 2a must equal m, which then tells us a = m/2 or a = .5m. The significance of c is its the value of y(t)=0. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Student Solution: If y = a t^2 + b t + c we have y ' (t) = 2 a t + b, which is equivalent to the given function y ' (t)=mt+b . Since 2at+b=mt+b for all possible values of t the parameter b is the same in both equations, which means that the coefficients 2a and m must be equal also. So if 2a=m then a=m/2. The depth function must therefore be y(t)=(1/2)mt^2+bt+c. c is not specified by this analysis, so at this point c is regarded as an arbitrary constant. c depends only on when we start our clock and the position from which the depth is being measured. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique rating #$&*ok ********************************************* Question: `q Explain why, given only the rate-of-depth-change function y' and a time interval, we can determine only the change in depth and not the actual depth at any time, whereas if we know the depth function y we can determine the rate-of-depth-change function y' for any time. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Becuase of the (c) variable, it tells us the start point. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Given the rate function y' we can find an approximate average rate over a given time interval by averaging initial and final rates. Unless the rate function is linear this estimate will not be equal to the average rate (there are rare exceptions for some functions over specific intervals, but even for these functions the statement holds over almost all intervals). Multiplying the average rate by the time interval we obtain the change in depth, but unless we know the original depth we have nothing to which to add the change in depth. So if all we know is the rate function, have no way to find the actual depth at any clock time. ANOTHER EXPLANATION: The average rate of change over a time interval is rAve = `dy / `dt. If we know rAve and `dt, then, we can easily find `dy, which is the change in depth. None of this tells us anything about the actual depth, only about the change in depth. If we don't know rAve but know the function r(t) we can't use the process above to get the exact change in depth over a given interval, though we can often make a pretty good guess at what the average rate is (for a quadratic depth function, as the quiz showed, you can actually be exact the average rate is just the rate at the midpoint of the interval; it's also the average of the initial and final rates; and all this is because for a quadratic the rate function is linear--if you think about those statements you see that they characterize a linear function, whose average on an interval occurs at a midpoint etc.). For anything but a linear rate function we can't so easily tell what the average is. However we do know that the rate function is the derivative of the depth function. So if we can find an antiderivative of the rate function, all we have to do to find the change in depth is find the difference in its values from the beginning to the end of the interval. This difference will be the same whichever antiderivative we find, because the only difference that can exist between two antiderivatives of a given rate function is a constant (whose derivative is zero). We have to develop some machinery to prove this rigorously but this is the essence of the Fundamental Theorem of Calculus. You might not understand it completely at this point, but keep coming back to this explanation every week or so and you will soon enough.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):Not sure after reading all this if I'm on the right track with what I said.
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Given Solution: ** Rate of depth change can be found from depth data. This is equivalent to differentiation. Given rate-of-change information it is possible to find depth changes but not actual depth. This is equivalent to integration. To find actual depths from rate of depth change would require knowledge of at least one actual depth at a known clock time. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique rating #$&*ok "