course Mth 271
Be sure you always include your access code. If I hadn't caught this, it would not have posted. Fortunately I did, but I'm very focused on the work and won't always catch such omissions.
Quiz 1
If the function y = .015 t2 + -1.7 t + 93 represents depth y vs. clock time t, then what is the average rate of depth change between clock time t = 13.9 and clock time t = 27.8? First I have to find the depth at each of these 2 times so I insert each time into the equation.
y = .015 (13.9)2 + -1.7 (13.9) + 93
y = .015 (193.21) + -23.63 + 93
y = 2.89815 + -23.63 + 93
y = 72.26815 at clock time 13.9
y = .015 (27.8)2 + -1.7 (27.8) + 93
y= .015 (772.84) + -47.26 + 93
y= 11.5926 + -47.26 + 93
y = 57.3326 at clock time 27.8
the average depth change between these 2 points is found by rise / run
(57.3326 - 72.26815) / (27.8 ?13.9) = -14.93555 / 13.9 = -1.0745
So the average rate of depth change between (13.9, 72.26815) and (27.8, 57.3326) is -1.0745
What is the rate of depth change at the clock time halfway between t = 13.9 and t = 27.8?
To find the rate of depth change at the clock time of 20.85 (which is half way between 13.9 and 27.8, I have to determine the formula for the rate of change at instantaneous time of t.
I start with my original formula of y(t) = .015 (t)2 + -1.7 (t) + 93 and insert the time of (t +Dt). The Dt stands for Delta t. (couldn? find a delta sign)
When typing we use `d to stand for the Delta sign.
Y(t +Dt)=.015 (t +Dt)2 + -1.7 (t +Dt) + 93
simplify to .015(t2 + 2tDt + Dt 2) -1.7t -1.7Dt +93
Y(t +Dt)=.015t2 + .015(2tDt) + .015 Dt 2 -1.7t -1.7Dt +93
Y(t +Dt)=.015t2 + .03tDt + .015 Dt 2 -1.7t -1.7Dt +93
then I need to subtract the equation Y(t) from Y(t +Dt)
Y(t +Dt)=.015t2 + .03tDt + .015 Dt 2 -1.7t -1.7Dt +93
Y(t) = .015 t2 + -1.7 t + 93
the .015 t2 , -1.7 t, and 93 cancel each other out to leave
.03tDt + .015 Dt 2 - 1.7Dt
then to find the slope I divide by Dt
(.03tDt + .015 Dt 2 - 1.7Dt) / Dt =.03t + .015Dt - 1.7
So as delta t approaches 0 Y1 (t) = .03t -1.7 then to find the slope at 20.35 I just insert 20.85 into the equation to get .03(20.85) -1.7 which equals a rate of -1.0745
Excellent development of this idea. Note that the midpoint rate is identical to the average rate for the entire interval. This is because the rate function, which you correctly derived, is linear. A linear function always takes its average value for an interval at the midpoint of that interval.
This also implies that, since only a quadratic function will give you a linear rate function, the property that the midpoint rate is equal to the average rate for the interval is unique to quadratic functions.
What function represents the rate r of depth change at clock time t?
Dy / Dt or [y(t) ?Y(t+Dt)] / [t ?(t+ Dt)]
What is the clock time halfway between t = 13.9 and t = 27.8, and what is the rate of depth change at this instant? Clock time is 20.85 rate of depth change at that time is -1.0745
If the function r(t) = .193 t + -2.1 represents the rate at which depth is changing at clock time t, then how much depth change will there be between clock times t = 13.9 and t = 27.8? would be the formula of
.193t + .0965Dt + (-2.1) which equals 4.60675
What function represents the depth? .193tDt + .0965 Dt2 + (-2.1Dt)
What would this function be if it was known that at clock time t = 0 the depth is 130 ? a quadratic equation. I? just at a loss on this question
A quadratic function y = a t^2 + b t + c yields the rate function r = 2 a t + b.
So if the rate function is r = m t + b, we identify m with 2 a and a = m / 2; b still identifies with b.
Thus the quadratic function is y = a t^2 + b t + c = (m/2) t^2 + b t + c.
In this case m = .193 and b = -2.1, so the quadratic function is
y = (.193/2) t^2 + (-2.1) * t + c, or
y = .0965 t^2 - 2.1 t + c.
Since c doesn't affect the rate function, it doesn't matter at all what c is. So unless we have other information we just leave c unspecified.
In this case, we do have information which can be used to find the specific value of c. Since at t = 0 the depth is y = 130, we substitute into our quadratic function to get
130 = .0965 * 0^2 - 2.1 * 0 + c, or just
c = 130.
The function is therefore
y = .0965 t^2 - 2.1 t + 130.