Question: fA particle moves such that its position is defined as s (t) = 40sin(5t) , t 20 14. The expression that represents the acceleration of

 \fA particle moves such that its position is defined as s(t) = 40sin(5t) , t 20 14. The expression that represents theacceleration of the particle is O 200cos (5t) O -1000sin (5t) O1000sin(5t) O -200cos(5t)15. The side of a square is increasing at arate of 5 cm/s. At what rate is the area changing whenthe side is 10 cm long? 0500m2/s 025cm2/s O100cm2/s 0200m2/s Math 31Formula Sheet PreCalculus Applications of Derivatives 43 + BB = (4+ B)42
- AB+ B2 ) $2 - 51 V2 - V1 V avga avg 43 - B' = (4-B)42 + AB+ B2) 12 -1 1 12 - 1 1 Triangle: Rectangle: If ax +bx+c=0, thenx=-bivb' -4ac =bh A =L . W 2a 2 P = sumof sides P = 2L + 2w m = 2 - V1y - y1 =m (x-x, ) Circle: Closed Box: (Rect. Prism) X2- X1 A = m-2 V =L . w. h D= V(x2

\fA particle moves such that its position is defined as s (t) = 40sin(5t) , t 20 14. The expression that represents the acceleration of the particle is O 200cos (5t) O -1000sin (5t) O 1000sin(5t) O -200cos(5t)15. The side of a square is increasing at a rate of 5 cm/s. At what rate is the area changing when the side is 10 cm long? 0500m2/s 025cm2/s O100cm2/s 0200m2/s Math 31 Formula Sheet PreCalculus Applications of Derivatives 43 + BB = (4+ B)42 - AB+ B2 ) $2 - 51 V2 - V1 V avg a avg 43 - B' = (4-B)42 + AB+ B2) 12 - 1 1 12 - 1 1 Triangle: Rectangle: If ax +bx+c=0, then x=-bivb' -4ac =bh A =L . W 2a 2 P = sum of sides P = 2L + 2w m = 2 - V1 y - y1 =m (x-x, ) Circle: Closed Box: (Rect. Prism) X2 - X1 A = m-2 V =L . w. h D= V(x2 -x, ) +(12 - V1) C = 2ar SA = 2L . w+ 2w.h+ 2L .h Sphere: (ball) Closed Cylinder: (can) V = arch Limits SA = 41 1-2 SA = 2xr2 + 2xrh lim f(x) - fla) lim flath)- fla) h-0 Right Angle Cone: x - a h V = arch a(r" -1) if r#1 r-1 SA = 2mrs where s = Vh2+12 S .. a if -1 0 sin x cos x - 1 lim = 0 1- cos x lim [ v(t ) at = s(1) + so Ja(t) at = v(1) + vo

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