קצבי שינוי מקשרים וגבולות מנה נשלפים
הוצג| English | משמעות |
|---|---|
| removable limit/rɪˈmuːvəbl ˈlɪmɪt/ | גבול הילך |
| related rate/rɪˈleɪtɪd reɪt/ | שיעור משתנה קשור |
A decision before an answer
- A water-depth rate is not a volume-flow rate: the tank’s cross-sectional area changes with depth. A quotient limit can likewise have a finite value without giving a differentiable extended function.
- Your goal: Translate a geometric rate into a derivative of the relevant quantity.
Read the relationship
- For a quantity V depending on a changing depth h(t), the chain rule gives dV/dt=V′(h) dh/dt. A draining tank has dh/dt<0, so signed dV/dt is negative; an outflow magnitude is its negative. Draw the geometry and name the instantaneous depth before substituting numbers. Do not treat the rate as a static volume divided by elapsed time unless the situation actually specifies a constant average rate.
- Derive and differentiate a spherical-cap volume formula.
A radius-3 spherical tank has depth 1 decreasing at 1/4 per time unit. What is the positive outflow magnitude?
Difficulty 1/5 · 100 coins + a small bonus on first completion today
Cross-section area is π(2Rh−h²)=5π. Multiply by depth-decrease magnitude 1/4.
Use the defining rule
- A spherical tank of radius R has cross-section radius squared R²−(R−h)²=2Rh−h² at depth h from the bottom. Integrating these circular areas gives the cap volume V(h)=π(Rh²−h³/3), for 0≤h≤2R. Its derivative π(2Rh−h²) is the current cross-sectional area. At R=3,h=1,dh/dt=−1/4, signed volume rate is −5π/4 and outflow magnitude 5π/4. The formula works for both shallow and deep caps within the stated range.
- Separate continuous quotient extension from differentiable extension.
For x≠0, f(x)=x|x|,g(x)=x. What happens when f/g is filled in at zero?
Difficulty 1/5 · 100 coins + a small bonus on first completion today
The quotient is |x|. Filling in zero gives continuity, while the left and right derivatives are −1 and 1.
Check the conditions
- If f and g are continuously differentiable near zero, f(0)=g(0)=0 and g′(0)≠0, then f(x)/g(x) tends to f′(0)/g′(0). This follows from f(x)=f′(0)x+o(x) and g(x)=g′(0)x+o(x); continuity of g′ keeps the quotient defined nearby except at zero. Filling in this limit gives a continuous extension. The argument does not require f′(0) nonzero, and it does not prove the extended quotient differentiable.
- Separate continuous quotient extension from differentiable extension.
For a sphere with R=3 and depth h=1, cross-sectional area is π(6−1)=5π. A depth decrease of 1/4 per time unit therefore removes volume at magnitude 5π/4 per time unit. For f(x)=sin(2x),g(x)=3x, the quotient extends at zero with value 2/3. This limit says nothing about the size of a tank and should not be substituted as a derivative rate without identifying the dependent quantities.
For f(x)=sin(2x) and g(x)=3x, express the removable quotient value as a fraction: ____.
Difficulty 1/5 · 100 coins + a small bonus on first completion today
The derivatives at zero are 2 and 3, so their quotient gives the limit 2/3.
Apply the task format
- For f(x)=x|x| and g(x)=x, both are continuously differentiable, f(0)=g(0)=0 and g′(0)=1. Their quotient for x≠0 is |x|, which extends continuously at zero but has unequal one-sided derivatives there. For (f²−f)/(2g−g³), factor to (f/g)(f−1)/(2−g²); its removable limit is −f′(0)/(2g′(0)). Additional factors approach −1 and 2. Distinguish the limit of the quotient from the derivative of its extension.
- Separate continuous quotient extension from differentiable extension.
Use instantaneous cross-section area, not total tank surface area. Report a signed volume change or a positive outflow as requested. A finite removable quotient limit establishes continuity, not differentiability of the filled-in quotient.
איזו תשובה מתאימה למקרה זה?
המרת קצב גיאומטרי לגזירה של הכמות הרלוונטית
A continuously differentiable numerator and denominator with simple denominator zero always give a differentiable quotient extension.
Difficulty 1/5 · 100 coins + a small bonus on first completion today
The numerator x|x| and denominator x meet the assumptions, but their filled-in quotient |x| is not differentiable at zero.
Keep the distinctions
- related rate 相关变化率 — A rate obtained by differentiating the relationship between changing quantities.
- removable limit 可去极限 — A finite nearby limit used to fill in a missing function value continuously.
- Translate a geometric rate into a derivative of the relevant quantity.
- Derive and differentiate a spherical-cap volume formula.
- Separate continuous quotient extension from differentiable extension.
Match each term with its precise meaning in this lesson.
Difficulty 2/5 · 140 coins + a small bonus on first completion today
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
Difficulty 3/5 · 200 coins + a small bonus on first completion today
The order follows the stated process; check each stage before the next.