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Q. 50

Expert-verifiedFound in: Page 860

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Annie is conscious of tidal currents when she is sea kayaking. This activity can be tricky in an area south-southwest of Cattle Point on San Juan Island in Washington State. Annie is planning a trip through that area and finds that the velocity of the current changes with time and can be expressed by the vector function

$>"\; separators="|">0.4\mathrm{cos}\left(\frac{\pi (t-8)}{6}\right),-1.1\mathrm{cos}\left(\frac{\pi (t-11)}{6}\right)$

where t is measured in hours after midnight, speeds are given in knots and point due north.

(a) What is the velocity of the current at 8:00 a.m.?

(b) What is the velocity of the current at 11:00 a.m.?

(c) Annie needs to paddle through here heading southeast, 135 degrees from north. She wants the current to push her. What is the best time for her to pass this point? (Hint: Find the dot product of the given vector function with a vector in the direction of Annie’s travel, and determine when the result is maximized.)

Ans:

(a) The velocity of the current at 8:00 a.m is $v\left(t\right)=\u27e80.4,0\u27e9$.

(b) The velocity of the current at 11:00am is $v\left(t\right)=\u27e80,-1.1\u27e9$

(c) The best time to pass her at the point is $11:00AM$

given,

$>"\; separators="|">0.4\mathrm{cos}\left(\frac{\pi (t-8)}{6}\right),-1.1\mathrm{cos}\left(\frac{\pi (t-11)}{6}\right)$

The velocity of current changes with time and can be expressed as

$v\left(t\right)=>"\; separators="|">0.4\mathrm{cos}\left(\frac{\pi (t-8)}{6}\right),-1.1\mathrm{cos}\left(\frac{\pi (t-11)}{6}\right)$

total time $t$ in 8 hours

Substitute $t=8$ in the expression of velocity.

localid="1649607446490" $\begin{array}{r}v\left(t\right)=>"\; separators="|">0.4\mathrm{cos}\left(\frac{\pi (8-8)}{6}\right),-1.1\mathrm{cos}\left(\frac{\pi (8-11)}{6}\right)\end{array}v\left(t\right)=>"\; separators="|">0.4\mathrm{cos}\left(0\right),-1.1\mathrm{cos}\left(-\frac{\pi}{2}\right)& $

Total time $t$ is $11$ hours.

Substitute $t=11$ in the expression of velocity.

$\begin{array}{r}v\left(t\right)=>"\; separators="|">0.4\mathrm{cos}\left(\frac{\pi (11-8)}{6}\right),-1.1\mathrm{cos}\left(\frac{\pi (11-11)}{6}\right)\end{array}v\left(t\right)=>"\; separators="|">0.4\mathrm{cos}\left(\frac{\pi}{2}\right),-1.1\mathrm{cos}\left(0\right)$

$r\left(t\right)=\u27e8\mathrm{cos}\left(45\right),\mathrm{sin}(-45)\u27e9$

For role="math" localid="1649607185923" $v\left(t\right)=\u27e80.4,0\u27e9$

$\begin{array}{r}v\left(t\right)\cdot r\left(t\right)=\u27e80.4,0\u27e9\cdot \u27e8\mathrm{cos}\left(45\right),\mathrm{sin}(-45)\u27e9\\ v\left(t\right)\cdot r\left(t\right)=0.4\mathrm{cos}\left(45\right)=0.2828\end{array}$

For $v\left(t\right)=\u27e80,-1.1\u27e9$

role="math" localid="1649607287767" $\begin{array}{r}v\left(t\right)\cdot r\left(t\right)=\u27e80,-1.1\u27e9\cdot \u27e8\mathrm{cos}\left(45\right),\mathrm{sin}(-45)\u27e9\\ v\left(t\right)\cdot r\left(t\right)=-1.1\mathrm{sin}(-45)=0.7778\\ v\left(t\right)\cdot r\left(t\right)\text{is maximum at 11:00 A.M.}\end{array}$

Therefore, the best time to pass her at this point is $11:00AM$.

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