Two Related Rate Problems?

r = 5

dV/dt = 3

V = π r² * h

V = 25π * h

dV/dt = 25π * dh/dt

dh/dt = (dV/dt) / 25π

dh/dt = 3 / (25π)

dh/dt = 0.0382

Height of water is increasing at rate of 0.0382 m/sec

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Let x be position of first car from start point

Let y be position of second car from start point

Let t be the time in hours

x = 60t . . . . . . dx/dt = 60

y = 25t . . . . . . dy/dt = 25

Let s be distance between the two cars

s² = x² + y²

At time t = 2 hours,

x = 120

y = 50

s² = 14400 + 2500 = 16900

s = 130

Differentiate with respect to t:

s² = x² + y²

2s ds/dt = 2x dx/dt + 2y dy/dt

s ds/dt = x dx/dt + y dy/dt

130 ds/dt = 120*60 + 50*25

130 ds/dt = 8450

ds/dt = 8450/130

ds/dt = 65

Distance between cars is increasing at a rate of 65 miles/hour

r = 5

V = pi*r^2*h = pi*5^2*h = 25*pi*h

Therefore, dV/dh = 25*pi

dV/dt = 3

By the Chain Rule:

dV/dt = dV/dh * dh/dt

3 = 25*pi * dh/dt

3/(25pi) = dh/dt

dh/dt =~ 0.038 meters per second

dh/dt =~ 38 mm per second

After t hours, first car is 60t miles south, and second car is 25t miles west

By Pythagoras, if the distance between them distance between them is d, we have:

d^2 = (25t)^2 + (60t)^2

d^2 = 625t^2 + 3600t^2

d^2 = 4225t^2

d = 65t

Rate of change of d (with respect to t):

dd/dt = 65

volume = πr²h

= 25πh m³

After 1 second the tank contains 3 m³.

3 = 25πh

h ≅ 0.038 m

The height is increasing at 0.038 m/sec.

a million. something is incorrect with this question. factor (5, -2) isn't on graph of x³y² = 200 (5,-2) is positioned on graph of x³y² = 500 (2,-5) is positioned on graph of x³y² = 200 Please verify question, and let us know in case you typed it in incorrect, or if the errors is on your e book. besides, to verify this subject, you may desire to differentiate implicitly with appreciate to time (t in minutes) 3x²y² dx/dt + 2x³y dy/dt = 0 Now replace x with it proper fee (5?) and y with its proper fee(-2?), and dx/dt with -4. Use those values to calculate dy/dt. it is your answer. be conscious: If equation of graph isn't x³y² = 200, then this could exchange results of by-product (until the only distinction is the consistent on top area) ______________________________ 2. obtainable crate: 4 thousand value in line with crate: $25 furnish changing at value of -0.3 thousand/day x = 4 p = 25 dx/dt = -0.3 px + 7x + 8p = 328 Differentiate with appreciate to time (t) x dp/dt + p dx/dt + 7 dx/dt + 8 dp/dt = 0 (x + 8) dp/dt + (p + 7) dx/dt = 0 (4 + 8) dp/dt + (25 + 7) (-0.3) = 0 12 dp/dt - 9.6 = 0 12 dp/dt = 9.6 dp/dt = 0.8 value is increasing at a value of $0.80 (80 cents) in line with day M?thm?m