{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

15.3Ex61-62

# 15.3Ex61-62 - 652 C H A P T E R 15 D I F F E R E N T I AT I...

This preview shows page 1. Sign up to view the full content.

652 C H A P T E R 15 DIFFERENTIATION IN SEVERAL VARIABLES (ET CHAPTER 14) SOLUTION The function ρ ( 33 , T ) is concave up (concave down) if ρ T ( 33 , T ) is an increasing (decreasing) function of T . We use Table 1 to estimate whether the function ρ T ( 33 , T ) is increasing or decreasing. We compute the following values: ρ T ( 33 , 2 ) ρ ( 33 , 4 ) ρ ( 33 , 2 ) 2 = 26 . 23 26 . 38 2 = − 0 . 075 ρ T ( 33 , 4 ) ρ ( 33 , 6 ) ρ ( 33 , 4 ) 2 = 26 26 . 23 2 = − 0 . 115 ρ T ( 33 , 6 ) ρ ( 33 , 8 ) ρ ( 33 , 6 ) 2 = 25 . 73 26 2 = − 0 . 135 ρ T ( 33 , 8 ) ρ ( 33 , 10 ) ρ ( 33 , 8 ) 2 = 25 . 42 25 . 73 2 = − 0 . 155 ρ T ( 33 , 10 ) ρ ( 33 , 12 ) ρ ( 33 , 10 ) 2 = 25 . 07 25 . 42 2 = − 0 . 175 These values indicate that ρ T ( 33 , T ) is a decreasing function of T , which means that the second derivative is negative, i.e., 2 ρ T 2 ( 33 , T ) < 0 and the graph of ρ ( 33 , T ) is concave down. 61. Compute f xyz for f ( x , y , z ) = sin ( yx ) + tan z + z 1 x x 1 Hint: Use a well-chosen order of differentiation on each term. SOLUTION At the points where the derivatives are continuous, the partial derivative f xyz may be performed in any order. To simplify the computation we first differentiate with respect to y . This gives f y ( x , y , z ) = y sin ( yx ) + 0 = cos ( yx ) y ( yx ) = x cos ( yx ) We now differentiate
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern