# Calculus 2 : Applications in Physics

## Example Questions

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### Example Question #1 : Integral Applications

Determine the length of the following function between

Explanation:

In order to begin the problem, we must first remember the formula for finding the arc length of a function along any given interval:

where ds is given by the equation below:

We can see from our equation for ds that we must find the derivative of our function, which in our case is dv/dt instead of dy/dx, so we begin by differentiating our function v(t) with respect to t:

Now we can plug this into the given equation to find ds:

Our last step is to plug our value for ds into the equation for arc length, which we can see only involves integrating ds. The interval along which the problem asks for the length of the function gives us our limits of integration, so we simply integrate ds from t=1 to t=4:

### Example Question #2 : Integral Applications

Find the coordinates of the center of mass for the region bounded by the following two functions:

Explanation:

Before we can set up any sort of integral, we must find out where our two functions intersect, which tells of what the bounds of our region are. To do this, we set our functions equal to each other and solve for the x values at which they intersect:

Our next step is to use the formulas for the x and y coordinates of any region's center of mass, which are given below:

We know the bounds of our region, as well as the functions f(x) and g(x) that make up its upper and lower boundaries, respectively, so the only thing we have left to do is calculate the area of our region using the following integral:

Now that we know the area, bounds, and functions f(x) and g(x) that make up our region, we can simply plug these values into the formulas for the x and y coordinates of the center of mass for the region:

So by evaluating our integrals, we can see that the center of mass of the region bounded by our two functions is   .

### Example Question #3 : Integral Applications

Calculate the surface area of the solid obtained by rotating

about the x-axis from  to .

Explanation:

Before we can begin the problem, we must remember the formula for finding the surface area of a solid formed from rotation about the x-axis:

where ds is given by the following equation:

So in order to set up our integral for surface area, we must first find ds, which requires us to evaluate the derivative of our function:

In order to add the two terms in our radicand, we must find their common denominator. Because one of our terms is just 1, we can simplify the equation by writing 1 as (16-x^2)/(16-x^2), which allows us to add the two terms together and gives us:

Now we have our expression for ds in terms of x, and we can write the y in the formula for surface area by replacing it with its equation in terms of x given in the problem statement:

### Example Question #1 : Applications In Physics

In physics, the work done on an object is equal to the integral of the force on that object dotted with its displacent.

This looks like  ( is work,  is force, and  is the infinitesimally small displacement vector). For a force whose direction is the line of motion, the equation becomes .

If the force on an object as a function of displacement is , what is the work as a function of displacement ? Assume  and the force is in the direction of the object's motion.

Not enough information

Explanation:

, so .

Both the terms of the force are power terms in the form , which have the integral , so the integral of the force is .

We know

.

This means

.

### Example Question #5 : Integral Applications

. Find if . Assume  is measured in meters and  is measured in seconds.

Explanation:

By the fundamental theorem of calculus, we know that

So,

evaluated from  to

### Example Question #181 : Integrals

The velocity of a rocket, in meters per second,  seconds after it was launched is modeled by

.

What is the total distance travelled by the rocket during the first four seconds of its launch?

Explanation:

The distance traveled by an object over an interval of time is the total area under the velocity curved during the interval. Therefore, we need to evaluate the definite integral

Rewriting expressions written using radical notation using fractional exponents can make applying the power rule to find the antiderivative easier.

Evaluating this integral using the power rule,

, we find:

### Example Question #7 : Integral Applications

A force of

is applied to an object. How much work is done, in Joules, moving the object from  to  meters?

Explanation:

The work done moving an object is the intergral of force applied as a function of distance,

.

To find the work done moving the object from x=1 to x=4, evaluate the definite interval using x=1 and x=4 as the limits of integration. Recall that the antiderivative of a polynomial is found using the power rule,

### Example Question #8 : Integral Applications

The velocity of an moviing object at time  is , and the position at time  is . The acceleration of the object is modeled by the function  for

Explanation:

The position function is the second antiderivative of acceleration. First, integrate the acceration function to find the velocity function. Since the acceleration function is a polynomial, the power rule is needed. Recall that the power rule is

.

.

Next, we can find the the value of the constant  using the fact that .

, so . The velocity function is therefore .

Now integrate the velocity function to find the position function.

.

Use the fact that  to find the constant

.

The velocity function is

### Example Question #1 : Initial Conditions

The temperature of an oven is increasing at a rate  degrees Fahrenheit per miniute for  minutes. The initial temperature of the oven is  degrees Fahrenheit.

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

Explanation:

Integrating  over an interval  will tell us the total accumulation, or change, in temperature over that interval. Therefore, we will need to evaluate the integral

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

A substitution is useful in this case. Let. We should also express the limits of integration in terms of . When , and when  Making these substitutions leads to the integral

.

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

In general,

.

Therefore,

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

degrees.

### Example Question #2 : Initial Conditions

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

and the velocity at time  of the particle is

Explanation:

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

We used the rule

to integrate.

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

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

C is therefore 30.

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

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