13. Dane:

𝑥 = 3𝑠𝑖𝑛𝑡

𝑦 = 2𝑐𝑜𝑠2

𝑡

Szukane:

𝑦 𝑥 , 𝑡

𝑥

0

𝑠𝑖𝑛𝑡 =

𝑥
3

sin

2

𝑡 =

𝑥

2

9

𝑐𝑜𝑠2𝑡 =

𝑦
2

𝑇𝑜ż𝑠𝑎𝑚𝑜ść: 𝑐𝑜𝑠2𝛼 = 2 cos

2

𝛼 − 1

2 cos

2

𝑡 =

𝑦
2

+ 1 cos

2

𝑡 =

𝑦
4

+

1
2

cos

2

𝑡 + sin

2

𝑡 = 1

𝑥

2

9

+

𝑦
4

+

1
2

= 1

𝑥

2

9

+

𝑦
4

=

1
2

𝑦
4

=

1
2

−

𝑥

2

9

𝑦 𝑥 =

4
2

−

4𝑥

2

9

= 2 −

4𝑥

2

9

𝑦 = 2𝑐𝑜𝑠2𝑡 0 = 2𝑐𝑜𝑠2𝑡
𝑐𝑜𝑠2𝑡 = 0
2 cos

2

𝑡 − 1 = 0

2 cos

2

𝑡 = 1

cos

2

𝑡 =

1
2

𝑐𝑜𝑠𝑡 =

2

2

𝑡 =

𝜋
4

14. 𝑥 = 𝐴𝑐𝑜𝑠 𝜔𝑡 𝑦 = 𝐵𝑠𝑖𝑛 𝜔𝑡 𝑐𝑜𝑠 𝜔𝑡 =

𝑥
𝐴

sin 𝜔𝑡 =

𝑦
𝐵

𝑥

2

𝐴

2

+

𝑦

2

𝐵

2

= 1

(uzupełnid)

15. 𝑣 𝑡 =

𝑑𝑥

𝑑𝑡

=

𝑏

𝑐

2

𝑐𝑡 + 𝑒

−𝑐𝑡

′

=

𝑏

𝑐

2

𝑐 − 𝑐𝑒

−𝑐𝑡

=

𝑏

𝑐

(1 − 𝑒

−𝑐𝑡

)

Zauważamy, że dla lim

𝑡→∞

𝑒

−𝑐𝑡

= 0, więc 𝑣

𝑚𝑎𝑥

=

𝑏

𝑐

∗ 1 − 0 = 𝑏/𝑐

Dla początku ruchu (gdy t=0), v jest równe 0.

𝑎 𝑡 =

𝑑𝑣

𝑑𝑡

=

𝑏

𝑐

′

−

𝑏

𝑐

∗ 𝑒

−𝑐𝑡

′

=

𝑏

𝑐

∗ 𝑐𝑒

−𝑐𝑡

= 𝑏 ∗ 𝑒

−𝑐𝑡

a będzie maksymalne, gdy 𝑒

−𝑐𝑡

będzie równe jeden, czyli t=0, więc 𝑎

𝑚𝑎𝑥

= 𝑏

16. 𝑥 =

𝑎

cos 𝜔𝑡

𝑦 = 𝑏 ∗ 𝑡𝑔 𝜔𝑡

cos 𝜔𝑡 =

𝑎
𝑥

𝑡𝑔 𝜔𝑡 =

𝑦
𝑏

=

𝑠𝑖𝑛 𝜔𝑡

𝑐𝑜𝑠 𝜔𝑡

𝑦
𝑏

∗

𝑎
𝑥

= sin 𝜔𝑡

sin

2

𝜔𝑡 + cos

2

𝜔𝑡 = 1

𝑦

2

𝑎

2

𝑏

2

𝑥

2

+

𝑎

2

𝑥

2

= 1

𝑦

2

𝑎

2

+ 𝑎

2

𝑏

2

𝑏

2

𝑥

2

= 1

𝑦

2

𝑎

2

= 𝑏

2

𝑥

2

− 𝑎

2

𝑏

2

= 𝑏

2

𝑥

2

− 𝑎

2

𝑦

2

=

𝑏

2

𝑎

2

𝑥

2

− 𝑎

2

𝑦 𝑥 =

𝑏
𝑎

𝑥

2

− 𝑎

2

𝑣

𝑥

𝑡 =

𝑑𝑥

𝑑𝑡

=

𝑎

cos 𝜔𝑡

′

=

𝑎

′

𝑐𝑜𝑠 𝜔𝑡 − 𝑎 ∗ 𝑐𝑜𝑠 𝜔𝑡

′

cos

2

𝜔𝑡

=

𝑎 ∗ 𝜔𝑡

′

∗ sin 𝜔𝑡

cos

2

𝜔𝑡

= 𝑎𝜔 ∗

𝑠𝑖𝑛 𝜔𝑡

cos

2

𝜔𝑡

=

𝑎𝜔𝑡𝑔 𝜔𝑡

𝑐𝑜𝑠 𝜔𝑡

𝑣

𝑥

𝑥 = 𝑎𝜔 ∗

𝑦
𝑏

∗

𝑥
𝑎

=

𝜔𝑥𝑦

𝑏

=

𝜔𝑥

𝑏

𝑎 𝑥

2

− 𝑎

2

𝑏

=

𝜔𝑥 𝑥

2

− 𝑎

2

𝑎

𝑣

𝑦

𝑡 =

𝑑𝑦

𝑑𝑡

= 𝑏𝑡𝑔 𝜔𝑡

′

=

𝑏𝜔

cos

2

𝜔𝑡

𝑣

𝑦

𝑦 =

𝑏𝜔𝑥

2

𝑎

2

𝑥

2

=

𝑦𝑎

𝑏

2

+ 𝑎

2

= 𝑎

2

𝑦

2

𝑏

2

+ 1

𝑣

𝑦

𝑦 = 𝑏𝜔

𝑦

2

𝑏

2

+ 1 =

𝑦

2

𝜔

𝑏

+ 𝑏𝜔

𝑣 𝑡 = (

𝑏𝜔

cos

2

𝜔𝑡

)

2

+ (

𝑎𝜔𝑠𝑖𝑛 𝜔𝑡

cos

2

𝜔𝑡

)

2

=

𝜔

2

𝑎

2

sin

2

𝜔𝑡 + 𝑏

2

𝜔

2

cos

4

𝜔𝑡

= 𝜔

2

(𝑎

2

sin

2

𝜔𝑡 + 𝑏

2

)/ cos

4

𝜔𝑡 =

𝜔

cos

2

𝜔𝑡

𝑎

2

sin

2

𝜔𝑡 + 𝑏

2

17.