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#### karush

##### Well-known member

- Jan 31, 2012

- 3,241

wasn't sure but made the \(\displaystyle \frac{da}{dt}\) negative since the ships are drawing closer to each other at least before 4pm

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- Thread starter
- #1

- Jan 31, 2012

- 3,241

wasn't sure but made the \(\displaystyle \frac{da}{dt}\) negative since the ships are drawing closer to each other at least before 4pm

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Your answer agrees with the result I got when I recently worked this problem when it was posted at Yahoo! Answers. This was the reply I gave:

Let's orient our coordinate axes such that ship A is at the origin at noon, and so ship B is at $(150,0)$. Thus, we may describe the position of each ship parametrically as follows:

Ship A:

\(\displaystyle x=35t\)

\(\displaystyle y=0\)

Ship B:

\(\displaystyle x=150\)

\(\displaystyle y=25t\)

If we let $D$ be the distance between the ships at time $t$, we may use the distance formula to write:

\(\displaystyle D^2(t)=(35t-150)^2+(25t)^2=50\left(37t^2-210t+450 \right)\)

Implicitly differentiating with respect to $t$, we find:

\(\displaystyle 2D(t)D'(t)=50(74t-210)=100(37t-105)\)

Hence:

\(\displaystyle D'(t)=\frac{50(37t-105)}{D(t)}=\frac{\sqrt{50}(37t-105)}{\sqrt{37t^2-210t+450}}\)

At 4:00 pm, we have $t=4$, and so we find:

\(\displaystyle D'(4)=\frac{\sqrt{50}(37(4)-105)}{\sqrt{37(4)^2-210(4)+450}}=\frac{215}{\sqrt{101}}\,\frac{\text{km}}{\text{hr}}\)

Thus, at 4:00 pm the distance between the ships is increasing at a rate of about 21.3933 kph.

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- #3

- Jan 31, 2012

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do you always use generally.

\(\displaystyle D^2(t)=x^2+y^2\)

with these triangle problems, it seems the book examples do the same. I guess its a nice template to use..

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- #4

Yes, for right triangles, it follows directly from the Pythagorean theorem.do you always use generally.

\(\displaystyle D^2(t)=x^2+y^2\)

with these triangle problems, it seems the book examples do the same. I guess its a nice template to use..