Question: MTH130 Module 2 - Section Assignment 3.6 (Homework) 1. Follow the steps below for the given function. (Do not use mixed numbers in your answers.)

 MTH130 Module 2 - Section Assignment 3.6 (Homework)1. Follow the stepsbelow for the given function. (Do not use mixed numbers in youranswers.) 2x + 7y = 7 Solve the equation for y. Differentiatethis equation with respect to x. V Complete the steps below toimplicitly take the derivative of the original equation. 2x + 7y =7 dy = 0 dx dy dx dy dxFollow the steps belowfor the given function. x3 - 7x2 - xy = 1 Solvethe equation for y. y Differentiate this equation with respect to x.V' Complete the steps below to implicitly take the derivative of theoriginal equation. (Leave your final answer in terms of x and y.)

MTH130 Module 2 - Section Assignment 3.6 (Homework)

1.

x5 - 7x2 - xy = 1 -xy = 1 - x3+ 7x2 d -xy) = d (1) d (x 3 ) .d (7x2 ) dx dx dx dx (V) - d (x) =0 - 3x2 + OX dx dy -X- = -3X2 + ydx dy dxFind BY by implicit differentiation, dx 2 9x + y2= 25 dy = dxDistance Between Two Cars A car leaves anintersection traveling west. Its position 4 sec later is 21 ft fromthe intersection. At the same time, another car leaves the same intersectionheading north so that its position 4 sec later is 26 ftfrom the intersection. If the speeds of the cars at that instant

Follow the steps below for the given function. (Do not use mixed numbers in your answers.) 2x + 7y = 7 Solve the equation for y. Differentiate this equation with respect to x. V Complete the steps below to implicitly take the derivative of the original equation. 2x + 7y = 7 dy = 0 dx dy dx dy dxFollow the steps below for the given function. x3 - 7x2 - xy = 1 Solve the equation for y. y Differentiate this equation with respect to x. V' Complete the steps below to implicitly take the derivative of the original equation. (Leave your final answer in terms of x and y.) x5 - 7x2 - xy = 1 -xy = 1 - x3 + 7x2 d -xy) = d (1) d (x 3 ) . d (7x2 ) dx dx dx dx (V) - d (x) = 0 - 3x2 + OX dx dy -X- = -3X2 + y dx dy dxFind BY by implicit differentiation, dx 2 9x + y2 = 25 dy = dxDistance Between Two Cars A car leaves an intersection traveling west. Its position 4 sec later is 21 ft from the intersection. At the same time, another car leaves the same intersection heading north so that its position 4 sec later is 26 ft from the intersection. If the speeds of the cars at that instant of time are 8 ft/sec and 14 ft/sec, respectively, find the rate at which the distance between the two cars is changing. (Round your answer to one decimal place.) ft/sEffect of Price on Supply of Eggs Suppose the wholesale price of a certain brand of medium-sized eggs p (in dollars/carton) is related to the weekly supply x (in thousands of cartons) by the following equation. 625p2 - x2 =100 If 40000 cartons of eggs are available at the beginning of a certain week and the price is falling at the rate of 8c/carton/week, at what rate is the supply changing? (Round your answer to the nearest whole number.) (Hint: To find the value of p when x = 40, solve the supply equation for p when x = 40.) cartons per weekVolume of a Cube The volume V of a cube with sides of length x in. is changing with respect to time. At a certain instant of time, the sides of the cube are 6 in. long and increasing at the rate of 0.3 in./s. How fast is the volume of the cube changing (in cu in/s) at that instant of time? cu in/sWatching a Helicopter Take Off At a distance of 44 ft from the pad, a man observes a helicopter taking off from a heliport. If the helicopter lifts off vertically and is rising at a speed of 51 ft/sec when it is at an altitude of 111 ft, how fast is the distance between the helicopter and the man changing at that instant? ( Round your answer to one decimal place.) ft/secFlow of Water from a Tank Water flows from a tank of constant cross-sectional area 80 ft through an orifice of constant cross-sectional area - ft located at the bottom of the tank. Initially, the height of the water in the tank was 20 ft, and t s later it was given by the equation 2vh + - 1 _t - 2V 20 = 0 0 s t = 80v 20. 40 How fast (in ft/s) was the height of the water decreasing when its height was 9 ft? ft/s

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