### All Calculus 3 Resources

## Example Questions

### Example Question #1 : Applications Of Partial Derivatives

Find the linear approximation to at .

**Possible Answers:**

**Correct answer:**

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 : Applications Of Partial Derivatives

Find the tangent plane to the function at the point .

**Possible Answers:**

**Correct answer:**

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:**

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 : Applications Of Partial Derivatives

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

**Possible Answers:**

**Correct answer:**

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 #5 : Applications Of Partial Derivatives

Find the equation of the plane tangent to at the point

**Possible Answers:**

**Correct answer:**

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 #6 : Applications Of Partial Derivatives

Find the equation of the tangent plane to at the point

**Possible Answers:**

**Correct answer:**

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 #7 : 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: ,

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 #3 : Lagrange's Theorem

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

**Possible Answers:**

is a maximum

is a minimum

There are no maximums or minimums

is a maximum

is a minimum

is a maximum

is a minimum

is a maximum

is a minimum

**Correct answer:**

is a maximum

is a minimum

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 #4 : Lagrange's Theorem

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

**Possible Answers:**

**Correct answer:**

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 #8 : Applications Of Partial Derivatives

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

**Possible Answers:**

**Correct answer:**

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