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Example Questions
Example Question #1 : Applications Of Partial Derivatives
Find the linear approximation to
at .
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 #1 : Applications Of Partial Derivatives
Find the tangent plane to the function
at the point .
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 #3 : Tangent Planes And Linear Approximations
Find the Linear Approximation to
at .None of the Above
We are just asking for the equation of the tangent plane:
Step 1: Find
Step 2: Take the partial derivative of
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 #3 : Tangent Planes And Linear Approximations
Find the equation of the plane tangent to
at the point .
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 : Tangent Planes And Linear Approximations
Find the equation of the plane tangent to
at the point
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 : Tangent Planes And Linear Approximations
Find the equation of the tangent plane to
at the point
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 , .Absolute Minimum:
Absolute Maximum:
Absolute Minimum:
Absolute Maximum:
Absolute Minimum:
Absolute Maximum:
Absolute Minimum:
Absolute Maximum:
,Absolute Minimum:
Absolute Maximum:
,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 #1 : Lagrange Multipliers
Find the minimum and maximum of
, subject to the constraint .is a maximum
is a minimum
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
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 : Lagrange Multipliers
Find the absolute minimum value of the function
subject to the constraint .
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 #3 : Lagrange Multipliers
Find the dimensions of a box with maximum volume such that the sum of its edges is
cm.
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