The pressure on an object P is proportional to the temperature and inversely proportional to the volume: 𝑃=𝑇𝑉

Question

The pressure on an object P is proportional to the temperature and inversely proportional to the volume: 𝑃=𝑇𝑉

P
=
T
V
with P measured in Pascals. In this case the temperature and the volume both depend on time, hence so the does the pressure: 𝑃(𝑡)=𝑇(𝑡)𝑉(𝑡)
P
(
t
)
=
T
(
t
)
V
(
t
)
. At time 𝑡=8
t
=
8
seconds the temperature of an object is 85𝐶
85
C
and increasing at a rate of −5𝐶𝑠
−
5
C
s
and the volume is 20𝑚3
20
m
3
and increasing at the rate −8𝑚2𝑠
−
8
m
2
s
.

Using the quotient rule, the pressure on the object is changing at the rate

Answer 1

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Answer 2

Well, the quotient rule states that if you have a function of the form f(t) = g(t) / h(t), then the derivative of f(t) is given by:

f'(t) = (g'(t) * h(t) - g(t) * h'(t)) / (h(t))^2

In this case, we have P(t) = T(t) * V(t), where P(t) is the pressure, T(t) is the temperature, and V(t) is the volume, both as functions of time.

So, let's differentiate P(t) using the quotient rule. We have:

P'(t) = (T'(t) * V(t) + T(t) * V'(t)) / (V(t))^2

Now, we can substitute the given values into the equation:

P'(8) = (T'(8) * V(8) + T(8) * V'(8)) / (V(8))^2

Since T'(t) is given to be -5°C/s and V'(t) is given to be -8 m²/s, and at t = 8, T(8) = 85°C and V(8) = 20 m³, we can fill in the values:

P'(8) = (-5 * 20 + 85 * -8) / (20)^2

Now, let me calculate that for you...

Calculating...

Alright, after some quick calculations, the rate at which the pressure is changing at time t = 8 seconds is approximately -21.25 Pa/s.

Please note that this answer is based on the given information and calculations, and there might be some rounding errors.

Answer 3

To find the rate at which the pressure on the object is changing, we use the quotient rule for differentiation.

The quotient rule states that if we have two functions, u(x) and v(x), then the derivative of their quotient is given by:

(d/dx)(u(x)/v(x)) = (v(x)*(du/dx) - u(x)*(dv/dx))/(v(x)^2)

In this case, we have P(t) = T(t) * V(t), where P(t) represents the pressure, T(t) represents the temperature, and V(t) represents the volume at time t.

So, T(t) and V(t) are functions of time, and we need to find dP/dt, the rate at which the pressure is changing with respect to time.

Applying the quotient rule, we have:

(dP/dt) = (V(t)*(dT/dt) - T(t)*(dV/dt))/(V(t)^2)

Given that at t = 8 seconds, T(8) = 85°C, dT/dt = -5°C/s, V(8) = 20 m^3, and dV/dt = -8 m^3/s, we can substitute these values into the equation:

(dP/dt) = (20*(-5) - 85*(-8))/(20^2)

Simplifying further:

(dP/dt) = (-100 + 680)/400

(dP/dt) = 580/400

(dP/dt) = 1.45 Pa/s

So, the pressure on the object is changing at a rate of 1.45 Pa/s.

Answer 4

To find the rate at which pressure is changing, we need to take the derivative of the pressure function with respect to time. In this case, since both temperature and volume are changing with respect to time, we will use the quotient rule to differentiate the pressure function.

The quotient rule states that if we have a function of the form f(t) = g(t) / h(t), where both g(t) and h(t) are differentiable, then the derivative of f(t) is given by:

f'(t) = (g'(t) * h(t) - g(t) * h'(t)) / [h(t)]²

In our case, g(t) is the temperature function T(t) and h(t) is the volume function V(t). Let's differentiate the pressure function P(t) = T(t) * V(t) using the quotient rule:

P'(t) = (T'(t) * V(t) - T(t) * V'(t)) / [V(t)]²

Given that at t = 8 seconds, T(t) = 85°C, T'(t) = -5°C/s, V(t) = 20 m³, and V'(t) = -8 m²/s, we can now substitute these values into the derivative formula to find the rate at which pressure is changing:

P'(8) = (T'(8) * V(8) - T(8) * V'(8)) / [V(8)]²

P'(8) = (-5 * 20 - 85 * (-8)) / [20]²

P'(8) = (-100 + 680) / 400

P'(8) = 580 / 400

P'(8) = 1.45 Pa/s

Therefore, the pressure on the object is changing at a rate of 1.45 Pa/s at time t = 8 seconds.