The 2003 SARS Outbreak In the early stages of the SARS (severe acute respiratory syndrome) epidemic in 2003, the number of reported cases could be approximated by A ( t ) = 167 ( 1.18 ) t ( 0 ≤ t ≤ 20 ) t days after March 17,2003 (the first day in which statistics were reported by the World Health Organization.) a. What, approximately, was the instantaneous rate of change of A ( t ) on March 27 ( t = 10 ) ? Interpret the result. b. Which of the following is true? For the first 20 days of the epidemic, the instantaneous rate of change of the number of cases (A) increased. (B) decreased. (C) increased and then decreased. (D) decreased and then increased.
The 2003 SARS Outbreak In the early stages of the SARS (severe acute respiratory syndrome) epidemic in 2003, the number of reported cases could be approximated by A ( t ) = 167 ( 1.18 ) t ( 0 ≤ t ≤ 20 ) t days after March 17,2003 (the first day in which statistics were reported by the World Health Organization.) a. What, approximately, was the instantaneous rate of change of A ( t ) on March 27 ( t = 10 ) ? Interpret the result. b. Which of the following is true? For the first 20 days of the epidemic, the instantaneous rate of change of the number of cases (A) increased. (B) decreased. (C) increased and then decreased. (D) decreased and then increased.
Solution Summary: The author calculates the instantaneous rate of change of the number of cases reported on March 27 using the balance difference quotient.
The 2003 SARS Outbreak In the early stages of the SARS (severe acute respiratory syndrome) epidemic in 2003, the number of reported cases could be approximated by
A
(
t
)
=
167
(
1.18
)
t
(
0
≤
t
≤
20
)
t days after March 17,2003 (the first day in which statistics were reported by the World Health Organization.)
a. What, approximately, was the instantaneous rate of change of
A
(
t
)
on March
27
(
t
=
10
)
? Interpret the result.
b. Which of the following is true? For the first 20 days of the epidemic, the instantaneous rate of change of the number of cases
Use Euler's method to numerically integrate
dy
dx
-2x+12x² - 20x +8.5
from x=0 to x=4 with a step size of 0.5. The initial condition at x=0 is y=1. Recall
that the exact solution is given by y = -0.5x+4x³- 10x² + 8.5x+1
Find an equation of the line tangent to the graph of f(x) = (5x-9)(x+4) at (2,6).
Find the point on the graph of the given function at which the slope of the tangent line is the given slope.
2
f(x)=8x²+4x-7; slope of the tangent line = -3
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