A large conical tank is positioned so that its vertex is pointed downward. Water drains out from a hole at the vertex. As the water drains, the height of the water (measured from the vertex to the top surface) is always twice the radius of the waters surface.
a. draw and label a diagram of the situation, defining any variables you use
b. write a function that gives the volume of the water in the cone as a function of radius of the water at the top surface, using no other variables/parameters.
c. using variable t to represent the time aspect of these changes, what rate of change notation can we use for how the volume of water changes with respect to time and how the radius of the water at the top surface changes with respect to time? (notation for the two quantities only, not an equation)
d. Starting from the original quantity relation in (b) determine the equation that relates their rates of change with respect to time, that is "find the related rates equation"
e.suppose that the water drains from the cone at a constant rate of 15 cm^3 per second. how fast is the radius of the top surface of the water changing when the radius measures 75 cm? write your result in a complete sentence
2. use implicit differentiation to determine dy/dx for the following relation: sin(2x^2y^3)-3x^3=1

Respuesta :

1. a) check the image attached

r= radius

h=height

b) Volume of conical tank ,V = (1/3)πr²h

= (1/3)πr²(2r)

V = 2/3πr³

c) Rate of change of volume with respect to time is denoted as dV/dt

Rate of change of radius is denoted as dr/dt

d) V = 2/3πr³

dV/dt =  2/3π (3r²) dr/dt

dV/dt = 2πr²  dr/dt

e) Rate of change in volume, dV/dt = 15cm³ / sec  ,  r =75cm

∴ 15 = 2π(75)² .  dr/dt

dr/dt = 15 / (2π(75)²)

dr/dt = 15/35342.92

dr/dt = 0.00042 cm/sec

Radius is changing at the rate of 0.00042cm/sec when the radius is 75cm.

2. sin(2x² y³) - 3x³ = 1

Differentiating :

cos(2x² y³) [2x² y³dy/dt + 2y³ 2x] - 9x² = 0

6x²y² dy/dx . cos(2x² y³) + 4xy³ cos(2x² y³) - 9x² = 0

dy/dx = 9x² - 4xy³ cos(2x² y³) / 6x²y² cos(2x² y³)

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