Calculus 1 : Equations

Study concepts, example questions & explanations for Calculus 1

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Example Questions

Example Question #81 : How To Find Solutions To Differential Equations

Which type of function solves the differential equation given.

Possible Answers:

Polynomial 

Rational

Logarithmic 

Exponential

Correct answer:

Polynomial 

Explanation:

Without doing very much, we can see that the solution to this differential equation will be such that the derivative is equal to the original function divided by 

One such type of function is a polynomial, where the derivative decreases the order of the function by , which is equivalent to dividing by .

We can also show this through integration.

First separate the variables.

Now take the integral remembering the rule for natural logs.

Thus we get,

From here recall the properties of natural logs such that .

Therefore we can rewrite our function and exponentiate it to solve for y.

.

From here, we can see that it is a polynomial.

Example Question #83 : How To Find Solutions To Differential Equations

Find the general solution to the differential equation:

Possible Answers:

Correct answer:

Explanation:

To find the general solution to the seperable differential equation, we must seperate the x and dx, y and dy to seperate sides:

Now, integrate both sides (note that the two constants of integration combine to make one C):

The following rules were used for integration:

Finally, solve for y:

 

Example Question #83 : How To Find Solutions To Differential Equations

Where will  have a local minimum or maximum between  and

Possible Answers:

,

Correct answer:

,

Explanation:

We must find the maximums and minimumss by setting the derivative equal to zero.

The derivative of  is found using the chain rule,

and the trigonometric rule of derivatives for cosine, 

.

Applying the above rules we find the derivative to be,

  when , so in this case, we must divide each angle by two since we are dealing with , so .

Example Question #91 : Differential Equations

Which one of the following functions is continuous and has continuous derivative at 2?

Possible Answers:

Correct answer:

Explanation:

In order for the graph to be continuous, both of the parts of the piecewise function must be equal.

For the derivative to be equal, one must take the derivatives at each part and see if this new piecwise function is equal at 2 as well.

Only

 

fufills both these requirements.

Remember to use the power rule to find the derivatives of the piecewise function, .

Example Question #91 : Solutions To Differential Equations

What is the derivative of   at ?

Possible Answers:

Correct answer:

Explanation:

Here you must use the chain rule,

and the rule for natural log,

.

First, you get 

,

then using the chain on , you get 

You must then remember the , which gives you

  as the final derivative.

Plugging in  gives .

Example Question #343 : Equations

Find  for the equation:

Possible Answers:

Correct answer:

Explanation:

Take the derivative of each term in the equation twice: with respect to  and then with respect to . When taking the derivative with respect to one variable, treat the other variable as a constant.

For the function

The derivative is then

Now bring  and  terms to opposite sides of the equation:

Now rearraging variables gives :

Example Question #2421 : Calculus

Find  for the following equation:

Possible Answers:

Correct answer:

Explanation:

Given the function:

We'll be taking the derivative with respect to the variables  and .

Beginning with the left side of the equation, only the  variable appears and the derivative is:

Note that 

Now for the right side of the equation, both  and  appear, so we'll utilize the chain rule giving the derivative:

Combining these gives the derivative of the original equation:

Since we're looking for , gather  terms on one side of the equation and  terms on the other side:

From there, separate terms once more to find :

 

Example Question #345 : Equations

Find  for the equation:

Possible Answers:

Correct answer:

Explanation:

For this problem, knowledge of the following derivatives is necessary:

To take the derivative of the equation

Let's begin with the left side. For each term, we'll take the derivative with respect to both variables  and , treating the other variable as just a constant when we do so. The derivative of the left side is thus

Now moving to the right side, the derivative is:

Notice how since the  term has no  term, when we take the derivative with respect to  we just get zero, since we're treating the  as a constant.

Now we have the derived equation:

Bring  and  terms to opposite sides of the equation:

Now we can once more rearrange variables to find :

 

Example Question #346 : Equations

Find  for the equation:

Possible Answers:

Correct answer:

Explanation:

We'll be taking the derivative of each term in the equation

with respect to both  and . When taking the derivative for one variable, treat the other variable as a constant:

Note that 

Notice how  is treated as a constant when taking the derivative with respect to  and so goes to zero.  The same happens when taking the derivative of  with respect to 

Now bring  and  terms to opposite sides of the equation:

Rearrange terms once more to find :

Example Question #347 : Equations

Find  for the equation:

Possible Answers:

Correct answer:

Explanation:

For this problem, note that:

To approach this problem, we'll take the derivative of each term twice: with respect to  and then with respect to . When taking the derivative with respect to one variable, treat the other variable as a constant.

For the equation

The derivative is then

Now, bring  and  terms to opposite sides of the equation:

Finally, rearrange terms once more to get :

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