Calculus 3 Review 1

    This topic is a collection of problems and concepts that might help someone understand their working knowledge of Calculus 3.

Show all work and justify each step.

(1) The graph of lies in a plane. What is the plane?

    Solution. To find the plane we let Thus and we notice that



Therefore, an equation of the tangent line is

(2) Determine so that the domain of the vector function is   where



  

    Solution. The function



has domain when since demands that Therefore, can be any real number less than

(3) Find the open intervals on which the curve given by the vector function



is smooth.

    Solution. First note that the vector function is continuous over any interval not containing Further,
    




only when both and .  Since the first equation has solution of and the second equation has solutions of and there are no real numbers such that   Therefore, is smooth over any open interval not containing   
    
(4) Find the indefinite integral .

    Solution. To find we ned to find both and Using integration by parts, we find
    


( with , and ) and applying integration by parts again we find



Thus,


Similarly, we find



Therefore,





where

(5) Describe and sketch the level curves for each of the following functions

        (a) given

        (b) given

    Solution. (a) We plot for as shown:



(b) We plot for as shown:



(6) State the domain for each of the following functions

        (a)          
    
        (b)  

    Solution. The domain of is and the domain of is



(7) Find the value of the following limits, if they exist.

    (a)      (b)

    Solution. (a) Along the path we find



Therefore, does not exist because the limit depends on the path. (b) Along the path we find



Therefore, does not exist because the limit depends on the path.

(8) Knowing that is a function of and , determine and   given

    Solution. Using implicit differentiation with respect to we find,   and so Using implicit differentiation with respect to we find,   and so

(9)  Find the equation of the tangent plane for each of the following surfaces at the given point.

        (a)   at the point
    
        (b) at

    Solution. (a) Let and then and So and Therefore the equation of the tangent plane at is which is
    (b) Using implicit differentiation with respect to we find, and so and Using implicit differentiation with respect to we find, and so and Therefore the equation of the tangent plane at is
    


which is
    
(10) Let Find a function such that  



    Solution. We compute the partial derivatives as:







Since,







we find satisfies the requirement.

(11) If show that

    Solution. We determine the partial derivatives as and   Therefore
    


(12) Find the points on the sphere where the tangent plane is parallel to the plane .

    Solution. Note that the plane can be written as



For the sphere, we determine and and the tangent plane we seek is parallel and so and where are the points we seek. Thus, and so yields



Solving for we have, Thus,



which yields the points and on the sphere where the tangent plane is parallel to the plane .  

(13) Find the total differential given

    Solution. We determine the partial derivatives as  





Therefore the total differential is


    
(14) Use differentials to approximate

    Solution. Let and we notice that is easy to compute We determine the total differential by computing the partial derivatives







Thus,






When,   the linear approximation formula yields









This approximation is accurate to one decimal place:      

• print this page
• pages linking here
• related pages

Cite this as:
Calculus 3 Review 1
Published by Library of Math -- Online math organized by subject into topics.
Written by Smith, David A.
http://www.libraryofmath.com/calculus-3-review-1.html