1 | initial version |

The maximum you need occurs in $[0,\pi/4]$. So, we can look for the maximum of the distance from the origin to the point over that interval as follows.

```
var('t')
ff = (exp(cos(t)) - 2*cos(4*t) + sin(t/12)^5)
v=vector((ff*cos(t),ff*sin(t)))
w=v[0]^2+v[1]^2
ans=w.find_maximum_on_interval(0,pi/4)
ans
```

which gives

```
(16.48247427444069, 0.74104978802589871)
```

So, the length is 16.4825.

The point at which this max occurs is given by:

```
v.subs(t=ans[1])
```

(2.99520450583, 2.74066127836)

2 | No.2 Revision |

The maximum you need occurs for $t$ in $[0,\pi/4]$. So, we can look for the maximum of the distance from the origin to the point over that interval as follows.

```
var('t')
ff = (exp(cos(t)) - 2*cos(4*t) + sin(t/12)^5)
v=vector((ff*cos(t),ff*sin(t)))
w=v[0]^2+v[1]^2
ans=w.find_maximum_on_interval(0,pi/4)
ans
```

which gives

```
(16.48247427444069, 0.74104978802589871)
```

So, the length is 16.4825.

The point at which this max occurs is given by:

```
v.subs(t=ans[1])
```

(2.99520450583, 2.74066127836)

3 | Fixed typo on length |

The maximum you need occurs for $t$ in $[0,\pi/4]$. So, we can look for the maximum of the distance from the origin to the point over that interval as follows.

```
var('t')
ff = (exp(cos(t)) - 2*cos(4*t) + sin(t/12)^5)
v=vector((ff*cos(t),ff*sin(t)))
w=v[0]^2+v[1]^2
ans=w.find_maximum_on_interval(0,pi/4)
ans
```

which gives

```
(16.48247427444069, 0.74104978802589871)
```

So, the length is ~~16.4825.~~$\sqrt{16.4825}=4.060$.

The point at which this max occurs is given by:

```
v.subs(t=ans[1])
```

(2.99520450583, 2.74066127836)

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