- Topic
- Related rates
- Difficulty
- Standard
- Suggested time
- 5 min
- Marks
- 3
Related rates practice question 14
A bead slides along a wire bent into the curve , with and in centimetres. Its -coordinate increases at 2 cm/s. At what rate is the bead’s distance from the origin changing as it passes the point ?
Give a signed rate in centimetres per second.
Only is given, yet the bead’s height changes as it slides. Which two relations hold at every instant: one for the distance to the origin, and one for the wire?
Let be the distance from the origin, so at every instant. Both and change with time.
The bead stays on the curve, so differentiate as well: .
A signed rate in cm/s, exact or to two decimal places.
Theory UBC Math 100 §5CLP-1 §3.2Key results
Worked solution
Strategy
The distance formula and the curve are two relations that hold at every instant. Differentiate both with respect to time, use the curve to find dy/dt, then substitute the point.
Working
Pythagoras from the origin to the bead, differentiated with respect to t.
Chain rule on y = 2√x, at x = 4: the y-coordinate changes too.
The distance at this instant.
Positive: the bead is moving away from the origin.
Answer
The answer is 3 over root 2, about 2.12 centimetres per second.
Check
Since y² = 4x, D = √(x² + 4x); 0.01 s later x = 4.02 and D ≈ 5.6781, up 0.0212 from 4√2 ≈ 5.6569.
Where this goes wrong
Treating y as a constant because only dx/dt is given, which gives √2. The curve ties y to x, so differentiate y = 2√x as well to find dy/dt.
Leans on
Implicit differentiation, from Implicit differentiation and inverse trigonometric functions
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