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Answer:
b) -9.8t +18
Step-by-step explanation:
b) The instantaneous rate of change is the derivative of the height function. The power rule is separately applied to each term, and the results added.
power rule: d/dx(x^n) = nx^(n-1)
Then the derivative is ...
dh/dt = 2(-4.9t^1) +18(1 ·t^0)
dh/dt = -9.8t +18
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If you haven't been introduced to derivatives as such, then you find the instantaneous rate of change by finding the average rate of change over a vanishing interval.
[tex]\dfrac{dh}{dt}=\displaystyle\lim\limits_{h\to 0}{ \dfrac{h(t+h)-h(t)}{h}}\\\\\dfrac{dh}{dt}=\lim\limits_{h\to 0}\dfrac{(-4.9(t+h)^2+18(t+h)+23)-(-4.9t^2+18t+23)}{h}\\\\\dfrac{dh}{dt}=\lim\limits_{h\to 0}\dfrac{-4.9(2th+h^2)+18h}{h}=\lim\limits_{h\to 0}{(-9.8t +18 -4.9h)}\\\\ \boxed{\dfrac{dh}{dt}=-9.8t +18}[/tex]
