# [SOLVED]distance between ships at 4 pm

#### karush

##### Well-known member
View attachment 1599

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

#### MarkFL

Staff member
Re: distance between ships at 4pm

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.

#### karush

##### Well-known member
Re: distance between ships at 4pm

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..

#### MarkFL

$$\displaystyle D^2(t)=x^2+y^2$$
Yes, for right triangles, it follows directly from the Pythagorean theorem. 