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 .
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To find the rate at which R is changing, we need to use the chain rule from calculus. The chain rule states that if a variable, in this case R, is dependent on two or more variables, we can find its rate of change with respect to another variable using the chain rule.
Given that V is increasing at a rate of 2 volts/sec and I is decreasing at a rate of 1/5 (amp/sec), we can write the equation as follows:
dV/dt = 2 volts/sec
dI/dt = -1/5 amp/sec
We also know the equation V = IR, so we can use implicit differentiation to find dR/dt:
Differentiating both sides of the equation V = IR with respect to time, we get:
dV/dt = d/dt(IR)
Using the product rule, we can differentiate the right side of the equation:
dV/dt = I(dR/dt) + R(dI/dt)
Now we substitute the given values of dV/dt and dI/dt:
2 = I(dR/dt) + R(-1/5)
To find the rate at which R is changing, we need to solve for dR/dt. We can rearrange the equation as follows:
I(dR/dt) = 2 + R(1/5)
Now substitute the given values of V, I, and R (V = 40 volts and I = 2 amp) into the equation:
2( dR/dt) = 2 + R(1/5)
2( dR/dt) = 2 + 40(1/5)
2( dR/dt) = 2 + 8
2( dR/dt) = 10
Finally, solve for dR/dt:
dR/dt = 10/2
dR/dt = 5
Therefore, the rate at which R is changing when V = 40 volts and I = 2 amp is 5 ohms/sec.