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AP Calculus BC : Initial Conditions

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Example Questions

Example Question #1 : Applications In Physics

The temperature of an oven is increasing at a rate $r(t)=2e^{0.5t}$ degrees Fahrenheit per miniute for $0\le t\le10$ minutes. The initial temperature of the oven is T(0)=75 degrees Fahrenheit.

What is the temperture of the oven at t = 5 ? Round your answer to the nearest tenth.

Possible Answers:

323.9

239.4

119.7

212.2

Correct answer:

119.7

Explanation:

Integrating r(t) over an interval [a,b] will tell us the total accumulation, or change, in temperature over that interval. Therefore, we will need to evaluate the integral

$\int_{0}^{5}2e^{0.5t}dt$

to find the change in temperature that occurs during the first five minutes.

A substitution is useful in this case. Let $u=0.5t, du=0.5dt, \ and\ 2du=dt$ . We should also express the limits of integration in terms of . When t = 0, u=0 , and when t=5, u=2.5. Making these substitutions leads to the integral

$\int_{0}^{2.5}4e^udu$ .

To evaluate this, you must know the antiderivative of an exponential function.

In general,

$\int e^xdx=e^x+c$ .

Therefore,

$\int_{0}^{2.5}4e^udu=4e^u|_{0}^{2.5}=4e^{2.5}-4\approx 44.7$ .

This tells us that the temperature rose by approximately 44.7 degrees during the first five minutes. The last step is to add the initial temperature, which tells us that the temperature at t=5 minutes is

75+44.7=119.7 degrees.

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Example Question #1 : Applications In Physics

Find the equation for the velocity of a particle if the acceleration of the particle is given by:

a(t)=16t^2+48

and the velocity at time t=0 of the particle is .

Possible Answers:

16t^2+48

16t^2+78

$\frac{16}{3}t^3+96$

$\frac{16}{3}t^3+48t+30$

Correct answer:

$\frac{16}{3}t^3+48t+30$

Explanation:

In order to find the velocity function, we must integrate the accleration function:

$\int (16t^2+48)dt=\frac{16}{3}t^3+48t+C$

We used the rule

$\int x^ndx=\frac{1}{n+1}x^{n+1}+C$

to integrate.

Now, we use the initial condition for the velocity function to solve for C. We were told that

v(0)=30

so we plug in zero into the velocity function and solve for C:

$v(0)=30=\frac{16}{3}(0)^3+48(0)+C$

C is therefore 30.

Finally, we write out the velocity function, with the integer replacing C:

$v(t)=\frac{16}{3}t^3+48t+30$

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Example Question #1 : Initial Conditions

Find the work done by gravity exerting an acceleration of $-10\frac{m}{s^2}$ for a 10kg block down from its original position with no initial velocity.

Remember that

$F_{grav}=mass*acceleration$

$W=\int_{a}^{b}F(x)dx$ , where F(x) is a force measured in Newtons , is work measured in Joules , and and are initial and final positions respectively.

Possible Answers:

1000J

-500J

100J

250J

Correct answer:

250J

Explanation:

The force of gravity is proportional to the mass of the object and acceleration of the object.

$F=ma= (10*-10)\frac{kg*m}{s^2}= -100N$

Since the block fell down 5 meters, its final position is and initial position is .

$\\ W=\int_{0}^{-5}-100dx\\ \\=\frac{-100x}{2}|_{0}^{-5} \\ \\=-50x|_{0}^{-5}\\ \\=-50(-5)--50(0)\\ \\=250J$

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Example Question #2 : Application Of Integrals

Evaluate the following integral and find the specific function which satisfies the given initial conditions:

$f(x)=\int3x^2-e^x+2dx$

F(0)=3

Possible Answers:

F(x)=x^3-e^x+2x+4

F(x)=x^3-e^x+2x-4

F(x)=x^3+e^x-2x+4

F(x)=x^3-e^x+2x

Correct answer:

F(x)=x^3-e^x+2x+4

Explanation:

Evaluate the following integral and find the specific function which satisfies the given initial conditions:

$f(x)=\int3x^2-e^x+2dx$

F(0)=3

To solve this problem,we need to evaluate the given integral, then solve for our constant of integration.

Let's begin by recalling the following integration rules:

$1) \int x^ndx=\frac{x^{n+1}}{n+1}+c$

$2) \int e^xdx=e^x+c$

Using these two, we can integrate f(x)

$f(x)=\int3x^2-e^x+2dx$

$\int3x^2-e^x+2dx=\frac{3x^3}{3}-e^x+2x+c=x^3-e^x+2x+c$

SO, we get:

F(x)=x^3-e^x+2x+c

We are almost there, but we need to find c. To do so, plug in our initial conditions and solve:

3=0^3-e^0+2(0)+c

3=-1+c

c=4

So, our answer is:

F(x)=x^3-e^x+2x+4

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AP Calculus BC : Initial Conditions

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #1 : Applications In Physics

Example Question #1 : Applications In Physics

Example Question #1 : Initial Conditions

Example Question #2 : Application Of Integrals

All AP Calculus BC Resources

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