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At noon, ship $A$ is $100$ km west of ship $B$. Ship $A$ is sailing east at $35$ km/hr and ship $B$ is sailing north at $25$km/hr. How fast is the distance between the ships changing at $4.00$pm.

1 Answer

Toolbox:
  • If $y=f(x)$ then $\large\frac{dy}{dx}$$=f'(x)$ is the rate of change of w.r.t x
  • $\large\frac{dy}{dx_{(x_1,y_1}}$ is the slope of the tangent to the curve at the point $(x_1,y_1) $ on the curve. It is the slope of the curve at that point.
  • The normal at a point $(x_1,y_1)$ on $y=f(x)$ is perpendicular to the tangent at $(x_1,y_1)$
  • It $y=f(x)$ and both depend on 't' then $\large\frac{dy}{dt}$$=f'(x).\large\frac{dx}{dt}$
Step 1:
Let the ships A and B be at the points Q and P at 12 noon (when t=0).
$PQ=100 km$
At time t, let the positions of the ships be Q' and P'.
The distance travelled by the ships would be 35t due east (ship A) and 25t due to north (ship B)
Therefore $PQ'=35t-100$ and
$PP'=25\;t$
The distance between the two ships at time t, is $P'Q'=s$
Step 2:
$s^2=(35t-100)^2+(25t)^2$
$\quad= (35t-100)^2+625\;t^2$ -----(i)
differentiating with respect to t,
$2s \large\frac{ds}{dt}$$=2(35t -100) \times 35 +1250 \;t$
$2s \large\frac{ds}{dt}$$=70(35t -100) +1250 \;t$
Step 3:
The rate of seperation at time t is
$\large\frac{ds}{dt}=\large\frac{1}{2s}$$\bigg[70 (35t-100)+1250\bigg]$
$\qquad=\large\frac{1}{s}$$\bigg[35 (35t-100)+625\;t\bigg]$-----(ii)
Step 4:
When the time is $4.00\;pm, t=4$
and the distance between them (from(i))
$s=\sqrt {(35 \times 4 -100)^2+(25 \times 4)^2}$
$\quad=\sqrt {1600+10000}$
$\quad=\sqrt {11600}$
$\quad= 20 \sqrt {29} km$
Step 5:
Therefore their rate of separation at this time (from (ii))
$\large\frac{ds}{dt}=\frac{1}{20 \sqrt {29}}$$(35(140-100)+2500)$
$\large\frac{ds}{dt}=\frac{1}{20 \sqrt {29}}$$(1400+2500)$
$\large\frac{ds}{dt}=\frac{3900}{20 \sqrt {29}}$
$\qquad=\large\frac{195}{\sqrt {29}}$$\;km/h$
answered Jul 23, 2013 by meena.p
 

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