When a point moves along multiple sides, the equation changes for each interval.
Example ①
In rectangle ABCD (AB=6, BC=4), point P moves A → B → C at 2 cm/s. Find the area y cm² of △APD x seconds after starting.
A→B（0≦x≦3）: AP=2x. y=(1/2)×2x×4=4x
B→C（3≦x≦5）: BP=2(x−3), PC=2x−6. The base of △APD is AB=6, and the shape of the triangle changes.
Answer: When P is on AB (0≦x≦3): y=4x. When P is on BC (3≦x≦5): the area calculation changes according to the situation.
In rectangle ABCD (AB=6, BC=4), point P moves A → B → C at 2 cm/s. Find the area y cm² of △APD x seconds after starting.
A→B（0≦x≦3）: AP=2x. y=(1/2)×2x×4=4x
B→C（3≦x≦5）: BP=2(x−3), PC=2x−6. The base of △APD is AB=6, and the shape of the triangle changes.
Answer: When P is on AB (0≦x≦3): y=4x. When P is on BC (3≦x≦5): the area calculation changes according to the situation.
Key Points
- The equation for the point's position changes for each interval.
- Check the value of x at interval boundaries (vertices).
- The graph often becomes a broken line.
Practice Problems
- In square ABCD (side length 6 cm), P moves A→B→C at 3 cm/s. When P is on AB (0≦x≦2), express the area y of △BPD in terms of x.
Answer and Explanation
- Answer: y=9x. BP=3x, and BD is a diagonal, so the height of the triangle is 6/2=3 (incorrect). Actually: when P is on AB, use the relationship between △APD and △BPD, such as y=half the area of the square－area of △APD. In class, perform the basic calculations.