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Serggg [28]
3 years ago
13

Linda is out jogging at a constant speed of 3 m/s. She jogs past Maggie, who is stationary but immediately takes off on her bicy

cle to try and catch up to Linda. Maggie accelerates on her bicycle at 2 m/s2. How long does it take Maggie to catch Linda
Mathematics
1 answer:
mina [271]3 years ago
7 0

Answer:

c

Step-by-step explanation:

just took it

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For the function y=3x2: (a) Find the average rate of change of y with respect to x over the interval [3,6]. (b) Find the instant
nirvana33 [79]

Answer:

The instantaneous rate of change of y with respect to x at the value x = 3 is 18.

Step-by-step explanation:

a) Geometrically speaking, the average rate of change of y with respect to x over the interval by definition of secant line:

r = \frac{y(b) -y(a)}{b-a} (1)

Where:

a, b - Lower and upper bounds of the interval.

y(a), y(b) - Function exaluated at lower and upper bounds of the interval.

If we know that y = 3\cdot x^{2}, a = 3 and b = 6, then the average rate of change of y with respect to x over the interval is:

r = \frac{3\cdot (6)^{2}-3\cdot (3)^{2}}{6-3}

r = 27

The average rate of change of y with respect to x over the interval [3,6] is 27.

b) The instantaneous rate of change can be determined by the following definition:

y' =  \lim_{h \to 0}\frac{y(x+h)-y(x)}{h} (2)

Where:

h - Change rate.

y(x), y(x+h) - Function evaluated at x and x+h.

If we know that x = 3 and y = 3\cdot x^{2}, then the instantaneous rate of change of y with respect to x is:

y' =  \lim_{h \to 0} \frac{3\cdot (x+h)^{2}-3\cdot x^{2}}{h}

y' =  3\cdot \lim_{h \to 0} \frac{(x+h)^{2}-x^{2}}{h}

y' = 3\cdot  \lim_{h \to 0} \frac{2\cdot h\cdot x +h^{2}}{h}

y' = 6\cdot  \lim_{h \to 0} x +3\cdot  \lim_{h \to 0} h

y' = 6\cdot x

y' = 6\cdot (3)

y' = 18

The instantaneous rate of change of y with respect to x at the value x = 3 is 18.

5 0
2 years ago
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