Calculus help! I know the answers but idk how to get them, pic below. I just need b,c,d
Second to last line, does that say \(h''(t)>0\) ?
yes
Using a linear approximation, you can show that \[h(t)\approx h'(a)(t-a)+h(a)\] for some known \(a\) close to a desired \(t\). In this case, you use \(a=5\) and \(t=4\). The last sentence tells you that \(h(a)=h(5)=6\), and the table says that \(h'(5)=0.7\). So, \[h(4)\approx 0.7(4-5)+6\] I'm not so sure about the "true value" part, though.
for part a it says the answer is 5.3 inches, less than true value
thank you:) any idea how to do parts b,c,d?
you have a macbook???
yes
I got to thinking about this problem a while ago, and the way to figure out the "true value" is by looking at the given values of \(h'(t)\). Since \(h''(t)>0\), you know that \(h'(t)\) must be increasing over the entire interval (as it's shown in the table). Judging by the values, you have \(h'(t)>0\) for the interval, which in turn means \(h(t)\) is increasing. So, since 4 comes before 5, the true value is indeed smaller than the estimated value.
yea i just need parts b,c,d
Right, sorry about that, didn't notice that in the description. The volume is given by \(V=lwh\), where \(h\) denotes the depth of water given by \(h(t)\), and the length \(l\) and width \(w\) of the container is a constant 10 and 20 (either respectively or vice versa). \(h\) is a function of time, \(h(t)\), so \(V=20\cdot10\cdot h(t)\). So, the rate of change of volume is given by \[\frac{dV}{dt}=200h'(t)\] (B) asks for \(\dfrac{dV}{dt}\) when \(t=2\), which is thus \[\frac{dV}{dt}\bigg|_{t=2}=200\cdot0.5=100\]
yay thats what it says in my answer booklet
Oh good, does the reasoning make sense?
yes
for part b how did you get 0.5 ?
To find \(\dfrac{dV}{dt}\) when \(t=2\), you have from the equation above, \[\frac{dV}{dt}=20\cdot10\cdot h'(2)\] \(h'(2)\) is given in the table.
ooh yeaa! lol my bad
any idea how to do c and d ?
Suppose this is a graph of \(h'(t)\) with the five desired "left-hand" rectangles drawn: |dw:1385525288290:dw|
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