Practice · Grade 11 Calculus
Catenary Curve — Practice
Graded drills on the catenary y = a·cosh(x/a) plus a boss question. Find the height y, check the sag and apply cosh with confidence. Grade 11.
Q1 of 7
0 correct
Height of the chain at its lowest point for a = 2:
y = 2 · cosh(0 ⁄ 2)
Quick answer
What's the best way to practise the catenary curve?
Work through problems in rising difficulty: always form the ratio x/a first, then evaluate cosh of it (cosh(t) = (eᵗ + e⁻ᵗ)/2) and finally multiply by a — this gives y = a·cosh(x/a). Check the lowest point (x = 0 → y = a) and make sure you never confuse cosh with the ordinary cos. Round to 4 decimal places like the calculator does and verify your work via the sag y − a.
HowTo
A 4-step solving strategy
This order works for every catenary problem — height y, sag, or comparison.
- 1Step 1 of 4
Read off a and x, check a > 0
Note the shape parameter a and the position x. The parameter a must be positive, otherwise x/a is not meaningfully defined.
- 2Step 2 of 4
Form the ratio x/a
Divide x by a — that is the argument of the hyperbolic cosine. Example: x = 3, a = 2 → x/a = 1.5.
- 3Step 3 of 4
Evaluate cosh and multiply by a
Compute cosh(x/a) = (e^(x/a) + e^−(x/a))/2 and multiply by a: y = a·cosh(x/a). Don't drop the leading factor a.
- 4Step 4 of 4
Round and check
Round to 4 decimal places like the calculator. Check: at x = 0 you must get y = a; the sag y − a is always ≥ 0.
Examples
Worked practice examples with full working
Four typical catenary problem types. Try each yourself first, then compare with the solution.
Easy
Height for a = 1, x = 1
y = 1 · cosh(1 ⁄ 1)
x/a = 1/1 = 1
cosh(1) ≈ 1.5431
y = 1 · 1.5431 = 1.5431
Check: cosh ≥ 1, so y ≥ a = 1 ✓
Base case with a = 1: here y equals cosh(x). Don't confuse it with cos(1) ≈ 0.5403.
Medium
Height for a = 2, x = 3
y = 2 · cosh(3 ⁄ 2)
x/a = 3/2 = 1.5
cosh(1.5) ≈ 2.3524
y = 2 · 2.3524 = 4.7048
Check: sag y − a = 4.7048 − 2 = 2.7048 ≥ 0 ✓
Standard problem. Ratio first, then cosh, then times a — in exactly that order.
Hard
Height for a = 3, x = 6
y = 3 · cosh(6 ⁄ 3)
x/a = 6/3 = 2
cosh(2) ≈ 3.7622
y = 3 · 3.7622 = 11.2866
Check: y − a = 11.2866 − 3 = 8.2866 ≥ 0 ✓
A larger ratio x/a = 2 makes y rise sharply — cosh grows fast for large arguments.
Hard
Boss: Height for a = 3, x = 7.5
y = 3 · cosh(7,5 ⁄ 3)
x/a = 7.5/3 = 2.5
cosh(2.5) ≈ 6.1323
y = 3 · 6.1323 = 18.3969
Check: y − a = 18.3969 − 3 = 15.3969 ≥ 0 ✓
A non-integer ratio (2.5) and a large argument — this shows how steeply cosh climbs. Round carefully.
Pitfalls
Common mistakes — and how to avoid them
These traps come up again and again in catenary problems.
Confusing cosh with cos
The catenary uses the hyperbolic cosine cosh, not the ordinary cos. cosh(1) ≈ 1.5431, but cos(1) ≈ 0.5403 — completely different values. cosh grows without bound and doesn't oscillate.
Forgetting the leading factor a
The formula is y = a·cosh(x/a), not just cosh(x/a). Compute cosh(x/a) first, then multiply by a — otherwise the scaling is missing.
Swapping x and a in x/a
The argument is x divided by a, i.e. x/a. For a = 2, x = 3 that is 1.5 — not a/x = 0.6667.
Rounding wrongly or too early
Carry enough digits through the calculation and round only the final result to 4 decimal places, exactly as the calculator shows it.
Treating the catenary as a parabola
y = a·cosh(x/a) is not a parabola y = c·x². Using the parabola formula gives different values — the two only roughly agree near x = 0.
Study
Practise with a plan — three quick tips
Memorise cosh values as anchors
cosh(0) = 1, cosh(1) ≈ 1.5431, cosh(2) ≈ 3.7622. With these reference values you'll spot rough arithmetic errors instantly.
Solve first, reveal the hint after
Write out your working before you look at the hint or solution. Active recall helps you learn far more than passive reading.
For every wrong answer: ask why
Was it cos instead of cosh? A missing factor a? A rounding slip? Note the cause — then you'll catch the mistake immediately next time.
FAQ
Frequently asked questions about practising
Glossary
Terms in one sentence
- Catenary
- The curve of a freely hanging chain: y = a·cosh(x/a).
- Parameter a
- The shape factor of the curve and also the height of its lowest point.
- cosh
- Hyperbolic cosine, cosh(t) = (eᵗ + e⁻ᵗ)/2.
- Argument
- The input value of cosh — here the ratio x/a.
- Lowest point
- The position x = 0, where y = a.
- Sag
- The rise above the lowest point, y − a.
- Boss question
- The last and hardest problem of a practice set.