Q231 · Practice questionTechnology-freeComplex unfamiliar9 marks
QUESTION 231 (9 marks)
A bead moves on . At (2,2), and .
a)[2 marks]
Determine dy/dx and dy/dt at that instant.
b)[4 marks]
Determine the second time derivative of y.
c)[3 marks]
Treat the moving point as a bead. At the same instant, determine the rate of change of its speed and decide whether it is speeding up. Use .
WORKED SOLUTION
9 marksPractice marking scheme
ANSWER
(a) ; . (b) . (c) units of speed per unit time; speeding up.
Worked solution
(a) Implicit differentiation gives . At (2,2), y'=-1. Chain rule gives .
(b) Differentiate spatially again:
At the point this gives . Then . Spatial and time second derivatives are different.
(c) Parts (a) and (b) give and . Their dot product is 9 and speed is . Consequently , so speed increases even though the y component of acceleration is negative.
Equivalent mathematically justified methods accepted.
Exact answers unless a decimal accuracy is specified.
Part a: Find the spatial gradient.
Part a: Apply the time chain rule.
Part b: Differentiate the spatial relation again.
Part b: Obtain spatial second derivative -1/3.
Part b: Apply the second time chain rule.
Part b: Calculate time acceleration -3.
Part c: Form both vector quantities from the earlier parts.
Part c: Evaluate the scalar-product speed derivative.
Part c: Interpret its positive sign.
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