University Physics II · topic UPII-06.02
Current with calculus
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Further practice
Work through a fresh case before tackling the extension.
Current with calculus: predict a finite change
The instantaneous current through a chosen surface is the time derivative of the net charge that has crossed it, i = dq/dt, with the sign fixed by which direction through the surface you called positive. Use the supplied topic model Y_out = i0*exp(-tc/tau), where Y_out is CURRENT i(t) (mA). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. tc is Cursor time t, measured in ms. Keep i0 (Initial current I₀) = 3 mA; tau (Time constant τ) = 2 ms fixed. For this exercise use only 1.50937 ms ≤ tc ≤ 8.99062 ms on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. Calculate Y_out at tc = 3.15525 ms and 4.20262 ms. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δtc across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
CURRENT i(t) against tc
Predictions from the supplied Current with calculus model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): CURRENT i(t) / mA
Horizontal axis (x): tc: Cursor time t / ms
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| tc: Cursor time t / ms | CURRENT i(t) / mA |
|---|---|
| 1.509375 | 1.4104725479993074 |
| 3.3796874999999997 | 0.5536450722634982 |
| 5.25 | 0.21731927110275437 |
| 7.120312499999999 | 0.08530314448487542 |
| 8.990625 | 0.033483576592556026 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of tc separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At tc = 3.15525 ms, substitute into Y_out = i0*exp(-tc/tau): Y₀ = 0.619395 mA.
- At tc = 4.20262 ms, the same fixed-parameter model gives Y₁ = 0.366887 mA.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.252507 mA; the input change is Δtc = 1.04737 ms.
- Average sensitivity is ΔY_out/Δtc = -0.241086 mA per ms. Its negative sign means that this response decreases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 0.619395 mA, Y₁ = 0.366887 mA; ΔY_out = -0.252507 mA; average sensitivity = -0.241086 mA per ms.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
Current with calculus: infer an input with uncertainty
The instantaneous current through a chosen surface is the time derivative of the net charge that has crossed it, i = dq/dt, with the sign fixed by which direction through the surface you called positive. Use the supplied topic model Y_out = i0*exp(-tc/tau), where Y_out is CURRENT i(t) (mA). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. tc is Cursor time t, measured in ms. Keep i0 (Initial current I₀) = 3 mA; tau (Time constant τ) = 2 ms fixed. For this exercise use only 1.50937 ms ≤ tc ≤ 8.99062 ms on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 0.14401232 mA, with a bounded reading uncertainty ±0.016579 mA. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown tc by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000748125 ms, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
CURRENT i(t) against tc
Predictions from the supplied Current with calculus model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): CURRENT i(t) / mA
Horizontal axis (x): tc: Cursor time t / ms
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| tc: Cursor time t / ms | CURRENT i(t) / mA |
|---|---|
| 1.509375 | 1.4104725479993074 |
| 3.3796874999999997 | 0.5536450722634982 |
| 5.25 | 0.21731927110275437 |
| 7.120312499999999 | 0.08530314448487542 |
| 8.990625 | 0.033483576592556026 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(tc) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(1.50937) = 1.41047 mA and Y_out(8.99062) = 0.0334836 mA. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives tc_line = 1.50937 + (0.144012 − 1.41047)(8.99062 − 1.50937)/(0.0334836 − 1.41047) = 8.39012 ms. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives tc = 6.07294 ms; substituting back gives Y_out = 0.14401232 mA.
- The interpolation error in the input is |tc_line − tc| = 2.31718 ms. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.12743332 mA and 0.16059132 mA. Sort the roots even if the response decreases: 5.85501 ms ≤ tc ≤ 6.31755 ms.
- The lower and upper input deviations are 0.217928 ms and 0.244611 ms. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer tc ≈ 6.07294 ms; compatible interval [5.85501, 6.31755] ms. Endpoint interpolation gives 8.39012 ms.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.