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Q1) Find the equation of the line tangent to the graph of f at (1, 11 ), where f

ID: 2841833 • Letter: Q

Question

Q1) Find the equation of the line tangent to the graph of f at (1, 11), where f is given by

f(x) = 9x3 ? 4x2 + 6

(Let x be the independent variable and y be the dependent variable.)






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Q3) What is the formula for V, the volume of a sphere of radius r?


V=



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Q4) The gravitational attraction, F, between the earth and a satellite of mass m at a distance r from the center of the earth is given by




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(b) Find the instantaneous velocity of the tomato at t = 1.8.


=                         sec OR sec/m OR m/sec OR m/sec^2 OR m



(c) What is the acceleration at t = 1.8?

=               m/sec2     



(d) How high does the tomato go?

=               m



(e) How long is the tomato in the air?

=                sec




Find the equation of the line tangent to the graph of f at (1, 11), where f is given by f(x) = 9x3 ? 4x2 + 6 (Let x be the independent variable and y be the dependent variable.) If f(x) = 6x3 + 9x2 ? 35x + 22, find the intervals on which f'(x) ? 1. What is the formula for V, the volume of a sphere of radius r? The gravitational attraction, F, between the earth and a satellite of mass m at a distance r from the center of the earth is given by At a time t seconds after it is thrown up in the air, a tomato is at a height of f(t) = ?4.9t2 + 25t + 4 meters. Round all answers below to two decimal places. What is the average velocity of the tomato during the first 1.8 seconds? (Enter the value for following and choose a unit for the first two)

Explanation / Answer

(1)


f(x) = 9x3 ? 4x2 + 6 at ( 1 ,11 )


slope = f ' = 27x^2 - 8x = 19


so equation is


y - 11 = 19 ( x - 1 )


=> y = 19x -19 + 11


=> y = 19x - 8



(2)


f(x) = 6x3+ 9x2? 35x + 22


f '(x) > = 1



=> 18x^2 + 18x - 35 > = 1


=> 18x^2 + 18x - 36 >= 0


=> (x-1)(x+2) > = 0


=> x belongs to ( - inf , -2 ] U ( 1 , inf )


or


x < = -2 and x > = 1



so option is



3




(3)


v = (4/3) pi r^3


dv / dr = (4/3) pi ( 3r^2 )


= 4 pi r^2


so it is surface area of sphere



(4)


F = GM m / r^2


dF/dr = -GMm / r^4 * 2r


= - 2 GMm / r^3



(5)


(a)


f(t) = ?4.9t2+ 25t + 4



avg velocity = f(1.8) - f(0) / 1.8 - 0


= 33.124-4 / 1.8


= 16.18 m/sec


(b)


f ' ( t ) = -9.8 t + 25


f ' ( 1.8 ) = 7.36 m/sec



(c)


accelaration = f " (t) = -9.8 m /sec^2



(d)


max height => f ' = 0


=> -9.8t + 25 = 0 =>t = 2.55


so max height is f ( 2.55) = 35.88 m