HW5 even solutions

# HW5 even solutions - Homework 5 Solutions 14.1.16 We have f...

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Unformatted text preview: Homework 5 Solutions 14.1.16 We have f t (18 , 6) ≈ Δ P Δ t = 90- 93 20- 18 =- 1 . 5 percent/month . Eighteen months after rates are exposed to a formaldehyde concentration of 6 ppm, the percent of rats surviving is decreasing at a rate of about 1.5 per month. In other words, during the eighteenth month, an additional 1.5% of the rats die. We have f c (18 , 6) ≈ Δ P Δ c = 82- 93 15- 6 =- 1 . 22 percent/ppm . If the original concentration increases by 1 ppm, the percent surviving after 18 months decreases by about 1.22. 14.1.18 (a) For points near the point (0 , 5 , 3) , moving in the positive x direction, the surface is sloping down and the function is decreasing. Thus, f x (0 , 5) < . (b) Moving in the positive y direction near this point the surface slopes up and the function increases, so f y (0 , 5) > . 14.2.2 We have f x ( x,y ) = 3 x 2 + 6 xy and f y ( x,y ) = 3 x 2- 4 y, so f x (1 , 2) = 15 and f y (1 , 2) =- 5 . 14.2.4 By the chain rule we have ∂z ∂x = ∂ ∂x £ ( x 2 + x- y ) 7 / = 7( x 2 + x- y ) 6 (2 x + 1) = (14 x 2 + 7)( x 2 + x- y ) 6 ∂z ∂y = ∂ ∂y £ ( x 2 + x- y ) 7 / = 7( x 2 + x- y ) 6 (- 1) =- 7( x 2 + x- y ) 6 14.2.6 V r = 2 3 πrh 14.2.8 ∂ ∂T 2 πr T ¶ =- 2 πr T 2 14.2.24 ∂ ∂M 2 πr 3 / 2 √ GM ¶ = 2 πr 3 / 2- 1 2 ¶ ( GM )- 3 / 2 ( G ) =- πr 3 / 2 M √ GM 14.2.26 z x = cos(5 x 3 y- 3 xy 2 )(15 x 2 y- 3 y 2 ) 1 14.2.30 ∂ ∂x 1 √ 2 πσ e- ( x- μ ) 2 / (2 σ 2 ) ¶ = 1 √ 2 πσ e- ( x- μ ) 2 / (2 σ 2 )- 2( x- μ ) 2 σ 2 ¶ 14.2.32 z x = 7 x 6 + yx y- 1 , and z y = 2 y ln2 + x y ln x 14.2.36 (a) The difference quotient for approximating f u ( u,v ) is given by f u ( u,v ) ≈ f ( u + h,v )- f ( u,v ) h . Putting ( u,v ) = (1 , 3) and h = 0 . 001 , the difference quotient becomes f u (1 , 3) ≈ 1 . 001(1 . 001 2 + 3 2 ) 3 / 2- 1(1 2 + 3 2 ) 3 / 2 . 001 ≈ 41 . 1 (b) By the product rule and chain rule, we have f u = ∂f ∂u = ( u 2 + v 2 ) 3 / 2 + u 3 2 ¶ ( u 2 + v 2 ) 1 / 2 (2 u ) = ( u 2 + v 2 ) 1 / 2 (4 u 2 + v 2 ) , so f u (1 , 3) = (1 2 + 3 2 ) 1 / 2 (4(1 2 ) + 3 2 ) ≈ 41 . 11 . We see that the approximation in part (a) was reasonable. 14.2.40 (a) To calculate ∂B ∂t , we hold P constant and differentiate with respect to t : ∂B ∂t = ∂ ∂t ( Pe rt ) = Pe rt r. In financial terms, ∂B ∂t represents the change in the amount of money in the bank as one unit of time passes by. (b) To calculate ∂B ∂P , we hold t constant and differentiate with respect to P : ∂B ∂P = ∂ ∂P ( Pe rt ) = e rt . In financial terms, ∂B ∂P represents the change in the amount of money in the bank at time t as you increase the amount of money that was initially deposited by one unit....
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HW5 even solutions - Homework 5 Solutions 14.1.16 We have f...

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