Calculus 3 : Applications of Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

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Example Questions

Example Question #21 : Applications Of Partial Derivatives

A box has a surface area of .  What length, width and height maximize the volume of the box?

Possible Answers:

Correct answer:

Explanation:

To optimize a function  subject to the constraint , we use the Lagrangian function, , where  is the Lagrangian multiplier.

If  is a three-dimensional function, the Lagrangian function expands to three equations,

 and .

 

In this problem, we are trying to maximize the volume of the box, so the equation being optimized is .  

The constraint is the surface area of the box, or .

 , 

Substituting these variables into the the Lagrangian function and the constraint equation gives us the following equations

We have four equations and four variables (,,  and ), so we can solve the system of equations.

Multiplying the first equation by  and the second equation by  gives us

The left side of both equations are the same, so we can set the right sides equal to each other

Multiplying the first equation by  and the second equation by  gives us

The left side of both equations are the same, so we can set the right sides equal to each other

 

We now know . Substituting  and  into the constraint gives us

 

These dimensions maximize the volume of the box.

 

Example Question #22 : Applications Of Partial Derivatives

A fish tank (right cylinder) with no top has a volume of . What height and radius will minimize the surface area of the fish tank?

Possible Answers:

 , 

 , 

 , 

 , 

Correct answer:

 , 

Explanation:

To optimize a function  subject to the constraint , we use the Lagrangian function, , where  is the Lagrangian multiplier.

If  is a two-dimensional function, the Lagrangian function expands to two equations,

 and .

 

In this problem, we are trying to minimize the surface area of the fish tank with no top, so the equation being optimized is .  

The constraint is the volume of the cylinder, or .

 , 

Substituting these variables into the the Lagrangian function and the constraint equation gives us the following equations

We have three equations and three variables (,, and ), so we can solve the system of equations.

Setting both expressions of lambda equal to each other gives us

Substituting this expression into the constraint, we have

 

These dimensions minimize the surface area of the fish tank.

 

Example Question #23 : Applications Of Partial Derivatives

A soda can (a right cylinder) has a volume of . What height and radius will minimize the surface area of the soda can?

Possible Answers:

Correct answer:

Explanation:

To optimize a function  subject to the constraint , we use the Lagrangian function, , where  is the Lagrangian multiplier.

If  is a two-dimensional function, the Lagrangian function expands to two equations,

 and .

 

In this problem, we are trying to minimize the surface area of the soda can, so the equation being optimized is .  

The constraint is the volume of the cylinder, or .

 , 

Substituting these variables into the the Lagrangian function and the constraint equation gives us the following equations

We have three equations and three variables (,, and ), so we can solve the system of equations.

Setting both expressions of lambda equal to each other gives us

Substituting this expression into the constraint, we have

 

These dimensions minimize the surface area of the soda can.

 

Example Question #24 : Applications Of Partial Derivatives

What is the least amount of fence required to make a yard bordered on one side by a house?  The area of the yard is .

Possible Answers:

Correct answer:

Explanation:

To optimize a function  subject to the constraint , we use the Lagrangian function, , where  is the Lagrangian multiplier.

If  is a two-dimensional function, the Lagrangian function expands to two equations,

 and .

 

In this problem, we are trying to minimize the perimeter of the yard, which is three sides, so the equation being optimized is .  

The constraint is the area of the fence, or .

 , 

Substituting these variables into the the Lagrangian function and the constraint equation gives us the following equations

We have three equations and three variables (,, and ), so we can solve the system of equations.

Setting the two expressions for  equal to each other gives us

Substituting this expression into the constraint gives us

These dimensions minimize the perimeter of the yard.

 

Example Question #25 : Applications Of Partial Derivatives

What is the least amount of wood required to make a rectangular sandbox whose area is ?

Possible Answers:

Correct answer:

Explanation:

To optimize a function  subject to the constraint , we use the Lagrangian function, , where  is the Lagrangian multiplier.

If  is a two-dimensional function, the Lagrangian function expands to two equations,

 and .

 

In this problem, we are trying to minimize the perimeter of the sandbox, so the equation being optimized is .  

The constraint is the area of the box, or .

 , 

Substituting these variables into the the Lagrangian function and the constraint equation gives us the following equations

We have three equations and three variables (,, and ), so we can solve the system of equations.

These dimensions minimize the perimeter of the sandbox.

 

Example Question #1 : Gradient Vector, Tangent Planes, And Normal Lines

Find the equation of the tangent plane to  at .

Possible Answers:

Correct answer:

Explanation:

First, we need to find the partial derivatives in respect to , and , and plug in .

 

Remember that the general equation for a tangent plane is as follows:

Now lets apply this to our problem

Example Question #27 : Applications Of Partial Derivatives

Find the slope of the function  at the point 

Possible Answers:

Correct answer:

Explanation:

To consider finding the slope, let's discuss the topic of the gradient.

For a function , the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

It is essentially the slope of a multi-dimensional function at any given point

Knowledge of the following derivative rules will be necessary:

Trigonometric derivative: 

Note that u may represent large functions, and not just individual variables!

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Looking at  at the point 

x:

y:

z:

 

Example Question #28 : Applications Of Partial Derivatives

Find the slope of the function  at the point 

Possible Answers:

Correct answer:

Explanation:

To consider finding the slope, let's discuss the topic of the gradient.

For a function , the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

It is essentially the slope of a multi-dimensional function at any given point

Knowledge of the following derivative rules will be necessary:

Trigonometric derivative: 

Note that u may represent large functions, and not just individual variables!

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Looking at  at the point 

x:

y:

z:

 

Example Question #29 : Applications Of Partial Derivatives

Find the slope of the function  at the point 

Possible Answers:

Correct answer:

Explanation:

To consider finding the slope, let's discuss the topic of the gradient.

For a function , the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

It is essentially the slope of a multi-dimensional function at any given point

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Looking at  at the point 

x:

y:

z:

 

Example Question #30 : Applications Of Partial Derivatives

Find the slope of the function  at the point 

Possible Answers:

Correct answer:

Explanation:

To consider finding the slope, let's discuss the topic of the gradient.

For a function , the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

It is essentially the slope of a multi-dimensional function at any given point.

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Looking at  at the point 

x:

y:

z:

 

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