AS 91578 · 2023 paper · gradient of normal, turning points, a proof
Question 3(c) Merit
Practises Parametric differentiation
A log chain rule, three readings off a drawn graph, a parametric normal on a Ferris wheel, stationary points of a reciprocal pair, and a catenary to verify by substitution.
Char rides a Ferris wheel. Her position is x = 5√2 sin((π t)/5) and y = 10 − 5√2 cos((π t)/5), where t is time in seconds from the start of the ride.
Find the gradient of the normal to this curve at the point when t = 6.25 seconds after the start of the ride.
| x | 5√2 sin((π t)/5) |
|---|---|
| y | 10 − 5√2 cos((π t)/5) |
| t | 6.25 s |
Work it out
Try it on paper first. Stuck? Take one hint at a time.
-
1 Differentiate x and y with respect to t. What does the chain rule bring out of (π t)/5?
A factor of π/5:
dx/dt = 5√2 · (π/5) cos(π t)/5 = √2πcos(π t)/5
dy/dt = 5√2 · (π/5) sin(π t)/5 = √2πsin(π t)/5 -
2 Divide.
dy/dx = (sin(π t/5))/(cos(π t/5)) = tan(π t)/5 -
3 At t = 6.25, what is the tangent’s gradient? Check your calculator is in radians.
6.25π/5 = 1.25π = 5π/4, and tan5π/4 = 1. -
4 Now the normal.
It is perpendicular, so its gradient is −1/1 = −1.
What the marker wanted
- Achieved Correct dx/dt and dy/dt.
- Merit The normal’s gradient, −1.
In our words, from NZQA’s 2023 marking schedule.
How did it go?
Got it moves it further out. Review brings it back tomorrow.