a) For this question, you can find the y-value at x = 4, then find the y-value at x = 4.2, and subtract to find Δy.
So, that's y = 3√4 = 3(2) = 6 for x = 4, and y = 3√4.2 = 3(2.049...) = 6.148...
Which leaves us with Δy = 6.148... – 6 ≈ 0.148
b) Here we apply the Power Rule, since a square root is the same as a one-half power.
This yields:
dy/dx = (3/2)x-1/2 = 3 ÷ (2√x)
And from there we can infer:
dy = (3 ÷ (2√x))dx
Into which we can simply plug in our x & dx values, which yields 3/20 or 0.15.
Rachel C.
are you willing to answer another question of mine? Let f (t) be the weight (in grams) of a solid sitting in a beaker of water. Suppose that the solid dissolves in such a way that the rate of change (in grams/minute) of the weight of the solid at any time 't' can be determined from the weight using the formula: f ' (t) = (-2)*(f(t))*(4 + f(t)). If there is 6 grams of solid at time t = 2 estimate the amount of solid 1 second later. ______________ grams
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11/02/23
Dillon W.
tutor
Sure.
We can use the values given to find the rate of change at t = 2.
f'(t) = -2•f(t)•(4+f(t)) = -8•f(t) – 2•f(t)•f(t)
Plugging in f(2) = 6, we get:
-8•6 – 2•6•6 = -48 – 72 = -120
Note that the problem specified this rate of change is in grams per minute.
So to find the rate in grams per second, we can divide by 60 and get -2 grams per second.
Then we take our 6 grams minus 2 grams for the second that elapsed to obtain our estimate, f(3) ≈ 4 grams.
Hope that was helpful, Rachel. If you need any more help, I am available for a lesson.
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11/02/23
Dillon W.
11/01/23