chapter7ProblemsAndSolutions

# chapter7ProblemsAndSolutions - Chapter 7 The Chain Rule...

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Unformatted text preview: Chapter 7 The Chain Rule, Related Rates, and Implicit Differentiation 7.1 For each of the following, find the derivative of y with respect to x . (a) y 6 + 3 y- 2 x- 7 x 3 = 0 (b) e y + 2 xy = √ 3 (c) y = x cos x Detailed Solution: (a) d dx ( y 6 + 3 y- 2 x- 7 x 3 ) = d dx (0) 6 y 5 dy dx + 3 dy dx- 2- 21 x 2 = 0 (6 y 5 + 3) dy dx = 21 x 2 + 2 dy dx = 21 x 2 + 2 6 y 5 + 3 (b) d dx ( e y + 2 xy ) = d dx ( √ 3) e y dy dx + (2 y + 2 x dy dx ) = 0 ( e y + 2 x ) dy dx =- 2 y dy dx =- 2 y e y + 2 x v.2005.1 - September 4, 2009 1 Math 102 Problems Chapter 7 (c) First take the logarithm of both sides: ln y = ln( x cos x ) = (cos x )(ln x ), then differentiate both sides with respect to x . d dx (ln y ) = d dx (cos x · ln x ) 1 y dy dx =- sin x · ln x + cos x · 1 x dy dx = y parenleftBig- sin x · ln x + cos x x parenrightBig = x cos x parenleftBig- sin x · ln x + cos x x parenrightBig 7.2 Consider the growth of a cell, assumed spherical in shape. Suppose that the radius of the cell increases at a constant rate per unit time. (Call the constant k , and assume that k > 0.) (a) At what rate would the volume, V , increase ? (b) At what rate would the surface area, S , increase ? (c) At what rate would the ratio of surface area to volume S/V change? Would this ratio increase or decrease as the cell grows? [Remark: note that the answers you give will be expressed in terms of the radius of the cell.] Detailed Solution: We are given the information that dr dt = k. (a) For the volume: V = 4 3 πr 3 dV dt = 4 π 3 d ( r 3 ) dt = 4 π 3 3 r 2 dr dt = 4 πr 2 k. (b) For the surface area: S = 4 πr 2 dS dt = d dt (4 πr 2 ) = 4 π (2 r ) dr dt = 8 πrk. (c) S V = 4 πr 2 (4 / 3) πr 3 = 3 r Thus d dt parenleftbigg S V parenrightbigg = d dt parenleftbigg 3 r parenrightbigg =- 3 r 2 dr dt =- 3 k r 2 . The derivative is negative so that the ratio of surface area to volume is decreasing as the cell grows. v.2005.1 - September 4, 2009 2 Math 102 Problems Chapter 7 7.3 Growth of a circular fungal colony A fungal colony grows on a flat surface starting with a single spore. The shape of the colony edge is circular (with the initial site of the spore at the center of the circle.) Suppose the radius of the colony increases at a constant rate per unit time. (Call this constant C .) (a) At what rate does the area covered by the colony change ? (b) The biomass of the colony is proportional to the area it occupies (factor of proportionality α ). At what rate does the biomass increase? Detailed Solution: The area of the colony is A = πr 2 and dr/dt = C . (a) dA/dt = π (2 r ) dr/dt = 2 πrC . (b) M = αA so dM/dt = αdA/dt = α 2 πrC . 7.4 Limb development During early development, the limb of a fetus increases in size, but has a constant proportion....
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chapter7ProblemsAndSolutions - Chapter 7 The Chain Rule...

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