DE's and Mathematical Modelling Quiz 1

DE's and Mathematical Modelling Quiz 1

12th Grade - University

10 Qs

quiz-placeholder

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DE's and Mathematical Modelling Quiz 1

DE's and Mathematical Modelling Quiz 1

Assessment

Quiz

Mathematics

12th Grade - University

Practice Problem

Medium

CCSS
HSF.BF.A.1, HSA.APR.D.6, HSA.APR.A.1

+6

Standards-aligned

Created by

Krista Edwards

Used 17+ times

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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Media Image

Shown above is a slope field for which differential equation?

dydx= xy\frac{dy}{dx}=\ xy

dydx=xy + y\frac{\text{d}y}{\text{d}x}=xy\ +\ y

dydx=xyy\frac{\text{d}y}{\text{d}x}=xy-y

dydx=xy+x\frac{\text{d}y}{\text{d}x}=xy+x

dydx=(x+1)3\frac{\text{d}y}{\text{d}x}=\left(x+1\right)^3

2.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

3.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

A rumour spreads among a population of N people at a rate proportional to the product of number of people who have heard the rumour and the number of people who have not heard the rumour. If p denotes the number of people who have heard the rumour, which of the following differential equations could be used to model this situation with respect to time t, where k is a positive constant?

4.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Media Image

Let y = f(t) be a solution to the differential equation  dydt=ky\frac{dy}{dt}=ky  , where k is a constant.  Values of f for selected values of t are given in the table above.  Which of the following is an expression for f(t)?

4t + 4

 2t2+42t^2+4  

 et2ln9+3e^{\frac{t}{2}\ln9}+3  

 4et2ln34e^{\frac{t}{2}\ln3}  

5.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

The population of the little town of Scorpion Gulch is now 1000 people. The population is presently growing at about 5% per year.

Find the general solution.

P=.05e1000t

P=1000e5t

P=1000e.05t

p=lne1000t

Tags

CCSS.HSF.BF.A.1

CCSS.HSF.LE.A.1

CCSS.HSF.LE.A.2

CCSS.HSF.LE.B.5

6.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

2. Solve the differential equation
 dydt=3t2y\frac{dy}{dt}=\frac{3t^2}{y}  with initial condition y(2) = 0.

𝑦 = ln(15t)

𝑦 =  16t316t^3  

 y=(2t316)12y=\left(2t^3-16\right)^{\frac{1}{2}}  

𝑦 =  (2t316)\left(2t^3-16\right)  

7.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

∫ 2x cos(x2) dx

sin( x2x^2 ) + C

2 sin ( 2x ) + C

12\frac{1}{2} sin( x2x^2 ) + C

4 cos(2x ) + C

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