It has been estimated that the number of Venutian Carp in the Tethis River is given by V(t)= 3t^2+8t+203 Where t is the number of years since 1990 A.Evaluate V'(5) B. Interpret what V'(5)means
What's stopping you?
what do you mean whats stopping me
Can't you do it on your own?
I am confussed on part A on where to begin
Do you know how to find \(V'(t)\)?
no
Well\[ V'(t) = \lim_{h\to 0}\frac{V(t+h)-V(t)}{h}\\ \]Which means: \[\begin{split} V'(5) &= \lim_{h\to 0}\frac{V(5+h)-V(5)}{h}\\ &=\lim_{h\to 0}\frac{3(5+h)^2+8(5+h)+203-[3(5)^2+8(5)+203]}{h} \end{split}\] Good luck =)
It would be really nice if you knew how to differentiate, wouldn't it?
yes
LOL wio giving first principle ><
Try to simplify that fraction a bit @farmergirl411
ok give me a minute
ok I got ((9h^2+168h+636/h))
is that right
No it is not.
what didi I do wrong
\[ \begin{split} V'(5) &= \lim_{h\to 0}\frac{V(5+h)-V(5)}{h}\\ &=\lim_{h\to 0}\frac{3(5+h)^2+8(5+h)+203-[3(5)^2+8(5)+203]}{h}\\ &=\lim_{h\to 0}\frac{3(5+h)^2+8(5+h)-[3(5)^2+8(5)]}{h}\\ &=\lim_{h\to 0}\frac{3(5)^2+3(10h)+3(h)^2+8(5)+8h-[3(5)^2+8(5)]}{h}\\ &=\lim_{h\to 0}\frac{3(10h)+3(h)^2+8h}{h}\\ &=\lim_{h\to 0}\frac{30h+3h^2+8h}{h}\\ &=\lim_{h\to 0}30+3h+8\\ &=30+8\\ &=38 \end{split} \]
I don't know where you went wrong, but wherever you didn't match me, you must have gone down the wrong path.
must be so then 38 is the anwer for part a
203−[3(5)2+8(5)+203] = -75-40 = -115
I am confussed who is right
im just showing u that u messed up the calculations for the constants
oh ok but so is the 38 right that wio got
6t+8= v'(5) = 38
yah
ok so you by any chance know how to do part b of this problem
you should practice simplying the first principle tho it usually cancels when you expand all the brackets!
ok
do you know how to do part b
Do you know that the first derivative gives you the slope or rate of change? In this case, the rate of change interpretation is more helpful. It tells you the rate of change in the number of carp
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