A presentation in Basic Calculus from Grade 11.

2. PROBLEM:
Car A leaves a parking lot at 9 am and
travels eastward at the rate of 60 kph. Another
vehicle, car B, leaves the same parking lot at 10
am and travels southward at the rate of 40
How fast is the distance between these two
are changing at 11 am?

4. (x)
9 am
60 Kph
(y)
10 am
40 Kph (z)
11 am
t=2

5. (x)
9 am
60 Kph
(y)
10 am
40 Kph (z)
11 am
t=2
1st Sentence:
𝒅𝒙
𝒅𝒕
= 𝟔𝟎 and 𝒙 𝒕 =
𝟔𝟎(𝒕)
2nd Sentence:
𝒅𝒚
𝒅𝒕
= 𝟒𝟎 and y 𝒕 − 𝟏 = 𝟒𝟎(𝒕 −
𝟏)

6. Find
𝒅𝒛
𝒅𝒕
at t=2
At t=2,
𝒙 𝟐 = 𝟔𝟎 𝟐
= 𝟏𝟐𝟎
𝒚 𝟐 − 𝟏 = 𝟒𝟎 𝟐 − 𝟏
= 𝟒𝟎
Pythagorean Theorem:
𝒙 𝟐
+ 𝒚 𝟐
= 𝒛 𝟐

7. 𝒛 𝟐 = 𝒙(𝒕) 𝟐 + 𝒚(𝒕 − 𝟏) 𝟐
= (𝟏𝟐𝟎) 𝟐+(𝟒𝟎) 𝟐
= 𝟏𝟒, 𝟒𝟎𝟎 + 𝟏, 𝟔𝟎𝟎
= 𝟏𝟔, 𝟎𝟎𝟎
≈ 𝟏𝟐𝟔. 𝟒𝟗𝟏𝟏𝟎𝟔𝟒

8. Pythagorean Theorem:
𝑥2 + 𝑦2 = 𝑧2
Derivative of Pythagorean Theorem:
2𝑥
𝑑𝑥
𝑑𝑡
+ 2𝑦
𝑑𝑦
𝑑𝑡
= 2𝑧
𝑑𝑧
𝑑𝑡

9. 2 120 60 + 2 40 40 = 2(126.4911064)
𝑑𝑧
𝑑𝑡
SUBSTITUTE
17600 = 252.9822128
𝑑𝑧
𝑑𝑡
17600
252.9822128
=
252.9822128
𝑑𝑧
𝑑𝑡
252.9822128

10. SUBSTITUTE
𝑑𝑧
𝑑𝑡
= 69.570 𝑘𝑝ℎ
𝑜𝑟
𝑑𝑧
𝑑𝑡
= 69.57 𝑘𝑝ℎ

11. CONCLUSIO
N
Therefore, the distance
between the two cars is
changing at 69.57 kph fast
at 11 am.