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1) If y = x3 + 2x and dx/ dt = 5, find dy/dt when x = 2. Give only the numerical answer. For example, if dy, dt = 3, type only 3.


2)The area A = πr2 of a circular puddle changes with the radius. At what rate does the area change with respect to the radius when r = 5ft?


5π ft2/ft

5 ft2/ft

10π ft2/ft

25π ft2/ft


3)A 25-ft ladder rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 0.18 ft/sec, how fast, in ft/sec, is the top of the ladder sliding down the wall, at the instant when the bottom of the ladder is 20 ft from the wall? Answer with 2 decimal places. Type your answer in the space below. If your answer is a number less than 1, place a leading "0" before the decimal point (ex: 0.35).

Respuesta :

Answer:

a) 70

b) 10π ft²/ft

c) 0.24 ft/sec

Step-by-step explanation:

1) y = x³ + 2x

[tex]\frac{dy}{dt}=\frac{d(x^3+2x)}{dt}[/tex]

or

[tex]\frac{dy}{dt}[/tex] = [tex]3x^2\frac{dx}{dt}+2\frac{dx}{dt}[/tex]

at [tex]\frac{dx}{dt}=5[/tex]  and x = 2

[tex]\frac{dy}{dt}[/tex] = [tex]3(2)^2\times(5)+2\times(5)[/tex]

or

[tex]\frac{dy}{dt}[/tex] = 60 + 10 = 70

2) A = πr² ft²

[tex]\frac{dA}{dr}=\frac{d(\pi r^2)}{dr}[/tex]

or

[tex]\frac{dA}{dr}[/tex]= 2(πr)

at r = 5 ft

[tex]\frac{dA}{dr}[/tex]= 2(π × 5) ft²/ft

or

[tex]\frac{dA}{dr}[/tex]= 10π ft²/ft

3) From Pythagoras theorem

Base² + Perpendicular² = Hypotenuse²

Thus,

B² + P² = H²  .............(1)

here, H = length of the ladder

P is the height of the wall

B is the distance from the wall at bottom

or

B² + P² = 25²       ...........(1)  

at B = 20 ft

20² + P² = 25²        

or

P² = 625 - 400

or

P = √225

or

P = 15 ft

differentiating (1) with respect to time, we get

[tex]2B\frac{dB}{dt}+2P\frac{dP}{dt}=0[/tex]

at B = 20 ft, [tex]\frac{dB}{dt} = 0.18[/tex] and P = 15 ft

⇒ 2(20)(0.18) + [tex]2(15)\frac{dP}{dt}[/tex] = 0

or

[tex]30\frac{dP}{dt}[/tex] = - 7.2

or

[tex]\frac{dP}{dt}[/tex] = - 0.24 ft/sec (Here negative sign depicts the ladder slides down)

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