The Voltage v , Current I and Resistance R are related by the equation V=IR. Suppose that V is increasing at a rate of 2 volt/sec, while I is decreasing of the rate of 1/5 (amp/sec ). Let t denote time in seconds.
Find the rate at which R is changing when V=40volts and 1=2 amp .
Well, let's put on our electrician's hat and solve this problem, shall we?
First, let's differentiate the equation V = IR with respect to time t:
dV/dt = dI/dt * R + I * dR/dt
The problem gives us dV/dt = 2 V/sec and dI/dt = -1/5 A/sec. We also know that when V = 40 V and I = 2 A.
Plugging in these values, we get:
2 = -1/5 * R + 2 * dR/dt
Simplifying, we have:
2 = -R/5 + 2 * dR/dt
Multiplying through by 5, we get:
10 = -R + 10 * dR/dt
Rearranging, we have:
10 * dR/dt = R - 10
Now, we know that when V = 40 V and I = 2 A, we can substitute these values to get:
40 = 2R
R = 20
Substituting this value of R back into our equation, we have:
10 * dR/dt = 20 - 10
10 * dR/dt = 10
Dividing both sides by 10, we get:
dR/dt = 1
So, the rate at which R is changing when V = 40 volts and I = 2 amps is 1 ohm/second.
Well, looks like R is changing at a steady pace. Let's hope it doesn't shock anyone!
To find the rate at which R is changing, we can use implicit differentiation.
Given that V = IR, we can differentiate both sides of the equation with respect to time t using the chain rule.
dV/dt = d(IR)/dt
Differentiating both sides with respect to t, we have:
dV/dt = I(dR/dt) + R(dI/dt)
Now, let's substitute the given values:
dV/dt = 2 volts/sec (as V is increasing at a rate of 2 volts/sec)
I = 2 amps (as I is 2 amps)
(dI/dt) = -1/5 amp/sec (as I is decreasing at a rate of 1/5 amp/sec)
By substituting these values into the equation, we have:
2 volts/sec = 2 amps (dR/dt) + R (-1/5 amp/sec)
Now, we need to find the rate at which R is changing (dR/dt) when V = 40 volts and I = 2 amps.
Substituting these values into the equation, we have:
2 volts/sec = 2 amps (dR/dt) + 40 (-1/5 amp/sec)
Simplifying further, we have:
2 = 2 (dR/dt) - 8
Rearranging the equation, we get:
2 (dR/dt) = 10
Dividing both sides by 2, we find:
dR/dt = 10/2
Therefore, the rate at which R is changing when V = 40 volts and I = 2 amps is 5 ohms/sec.
To find the rate at which R is changing, we can use implicit differentiation. Since V = IR, we can differentiate both sides of the equation with respect to time, t.
dV/dt = d/dt(IR)
Using the product rule, we can rewrite this as:
dV/dt = I(dR/dt) + R(dI/dt)
Given that dV/dt = 2 volts/sec and dI/dt = -1/5 amp/sec, and we are looking for dR/dt when V = 40 volts and I = 2 amps, we can substitute these values into the equation:
2 = (2)(dR/dt) + (40)(-1/5)
Simplifying this equation, we get:
2 = 2(dR/dt) - 8
Rearranging the equation, we have:
2(dR/dt) = 2 + 8
2(dR/dt) = 10
Finally, solving for dR/dt, we divide both sides of the equation by 2:
dR/dt = 10/2
dR/dt = 5
Therefore, the rate at which R is changing when V = 40 volts and I = 2 amps is 5 ohms/sec.
V = IR
dV/dt = R dI/dt + I dR/dt
Now just plug in your numbers.