Calculus 3 : Applications of Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

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Example Questions

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Example Question #1 : Applications Of Partial Derivatives

Find the linear approximation to  at .

Possible Answers:

Correct answer:

Explanation:

The question is really asking for a tangent plane, so lets first find partial derivatives and then plug in the point.

Remember that we need to build the linear approximation general equation which is as follows.

Example Question #2 : Tangent Planes And Linear Approximations

Find the tangent plane to the function  at the point .

Possible Answers:

Correct answer:

Explanation:

To find the equation of the tangent plane, we use the formula

.

Taking partial derivatives, we have

Substituting our values into these, we get

Substituting our point into , and partial derivative values in the formula we get

.

 

Example Question #1 : Applications Of Partial Derivatives

Find the Linear Approximation to  at .

Possible Answers:

None of the Above

Correct answer:

Explanation:

We are just asking for the equation of the tangent plane:

Step 1: Find 



Step 2: Take the partial derivative of  with respect with (x,y):



Step 3: Evaluate the partial derivative of x at 



Step 4: Take the partial derivative of  with respect to :



Step 5: Evaluate the partial derivative at 

.

Step 6: Convert (x,y) back into binomials:




Step 7: Write the equation of the tangent line:

Example Question #4 : Tangent Planes And Linear Approximations

Find the equation of the plane tangent to  at the point .

Possible Answers:

Correct answer:

Explanation:

To find the equation of the tangent plane, we find:  and evaluate  at the point given. , and . Evaluating  at the point  gets us . We then plug these values into the formula for the tangent plane: . We then get . The equation of the plane then becomes, through algebra, 

Example Question #1 : Tangent Planes And Linear Approximations

Find the equation of the plane tangent to  at the point 

Possible Answers:

Correct answer:

Explanation:

To find the equation of the tangent plane, we find:  and evaluate  at the point given. , and . Evaluating  at the point  gets us . We then plug these values into the formula for the tangent plane: . We then get . The equation of the plane then becomes, through algebra, 

Example Question #1 : Applications Of Partial Derivatives

Find the equation of the tangent plane to  at the point 

Possible Answers:

Correct answer:

Explanation:

To find the equation of the tangent plane, we need 5 things:

Using the equation of the tangent plane

, we get

Through algebraic manipulation to get z by itself, we get

Example Question #1 : Applications Of Partial Derivatives

Find the absolute minimums and maximums of  on the disk of radius .

Possible Answers:

Absolute Minimum: 

Absolute Maximum:

Absolute Minimum: 

Absolute Maximum: 

Absolute Minimum: 

Absolute Maximum: 

Absolute Minimum: 

Absolute Maximum: 

Absolute Minimum: 

Absolute Maximum:  

Correct answer:

Absolute Minimum: 

Absolute Maximum: 

Explanation:

The first thing we need to do is find the partial derivative in respect to , and .

 

We need to find the critical points, so we set each of the partials equal to .

We only have one critical point at , now we need to find the function value in order to see if it is inside or outside the disk.

This is within our disk.

 

We now need to take a look at the boundary, . We can solve for , and plug it into .

We will need to find the absolute extrema of this function on the range . We need to find the critical points of this function.

The function value at the critical points and end points are:

Now we need to figure out the values of  these correspond to.

Now lets summarize our results as follows:

 

From this we can conclude that there is an absolute minimum at , and two absolute maximums at  and .

Example Question #1 : Abstract Algebra

Find the minimum and maximum of , subject to the constraint .

Possible Answers:

 is a maximum

 is a minimum

 is a maximum

 is a minimum

 is a maximum

 is a minimum

 

There are no maximums or minimums

 is a maximum

 is a minimum

Correct answer:

 is a maximum

 is a minimum

Explanation:

First we need to set up our system of equations.

Now lets plug in these constraints.

 

Now we solve for 

If

 

 

If

 

 

Now lets plug in these values of , and  into the original equation.

 

We can conclude from this that  is a maximum, and  is a minimum.

Example Question #2 : Abstract Algebra

Find the absolute minimum value of the function  subject to the constraint .

Possible Answers:

Correct answer:

Explanation:

Let To find the absolute minimum value, we must solve the system of equations given by

.

So this system of equations is

, , .

Taking partial derivatives and substituting as indicated, this becomes

.

From the left equation, we see either or . If , then substituting this into the other equations, we can solve for , and get , , giving two extreme candidate points at .

On the other hand, if instead , this forces from the 2nd equation, and from the 3rd equation. This gives us two more extreme candidate points; .

 

Taking all four of our found points, and plugging them back into , we have

.

Hence the absolute minimum value is .

 

Example Question #1 : Lagrange Multipliers

Find the dimensions of a box with maximum volume such that the sum of its edges is  cm.

Possible Answers:

Correct answer:

Explanation:

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