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**Craka** · September 10th 2008, 05:25 PM

"An angler has a fish on the end of the line which is reeled in at 20cm/s from a bridge 3m above the water. Assuming that the fish is on the surface of the water, at what speed is the fish moving through the water when the amount of line out is 4m?"

Answer is given as 0.25m/s

I'm not sure how to even start this question. I would have thought the speed was the same as that being reeled in at of 0.2m/s???

**hatsoff** · September 10th 2008, 05:38 PM

The fishing rod is 3m above the water, which means that this is a trig problem. We have a right triangle with the length of the line as the hypotenuse, and the surface of the water from the foot of the bridge to the fish and the 3m height of the line as the catheti (the other two sides).

**symstar** · September 10th 2008, 06:06 PM

This forms a right triangle where we know the lengths of the height and hypotenuse.

The height is 3m and the hypotenuse is 4m. We can find the base to be $\sqrt{7}$ m with the Pythagorean Theorem.

Both the base and hypotenuse are changing with time, thus we could call the base x'(t) and the hypotenuse z'(t). We already know that z'(t)=0.2m/s.

You might try implicitly differentiating the Pythagorean Theorem with respect to time: $x^2+3^2=z^2$

**Craka** · September 10th 2008, 06:10 PM

still a little lost on how to solve

**skeeter** · September 10th 2008, 07:07 PM

as symstar said ...

$\frac{d}{dt} (x^2 + 3^2 = z^2)$

$2x \frac{dx}{dt} = 2z \frac{dz}{dt}$

$\frac{dx}{dt} = \frac{z}{x} \cdot \frac{dz}{dt} $

you were given the value of z, dz/dt, and you can use Pythagoras to calculate x ... now determine the value of dx/dt, the speed of the fish through the water.

**Craka** · September 10th 2008, 07:16 PM

This is what I get
$ \begin{array}{l} 2z(\frac{{dz}}{{dt}}) = 2x(\frac{{dx}}{{dt}}) \\ 2z \times 0.2 = 2x(\frac{{dx}}{{dt}}) \\ 0.4z = 2x(\frac{{dx}}{{dt}}) \\ \frac{{dx}}{{dt}} = \frac{{0.4z}}{{2x}} \\ \end{array} $

When calculating

$ \begin{array}{l} z = 4 \\ \frac{{dz}}{{dt}} = 0.2 \\ x = \sqrt 7 \\ \frac{{dx}}{{dt}} = \frac{4}{{\sqrt 7 }} \times 0.2 = 0.302 \\ \end{array} $

However the answer is given as 0.25

**symstar** · September 10th 2008, 11:07 PM

Originally Posted by Craka

This is what I get
$ \begin{array}{l} 2z(\frac{{dz}}{{dt}}) = 2x(\frac{{dx}}{{dt}}) \\ 2z \times 0.2 = 2x(\frac{{dx}}{{dt}}) \\ 0.4z = 2x(\frac{{dx}}{{dt}}) \\ \frac{{dx}}{{dt}} = \frac{{0.4z}}{{2x}} \\ \end{array} $

When calculating

$ \begin{array}{l} z = 4 \\ \frac{{dz}}{{dt}} = 0.2 \\ x = \sqrt 7 \\ \frac{{dx}}{{dt}} = \frac{4}{{\sqrt 7 }} \times 0.2 = 0.302 \\ \end{array} $

However the answer is given as 0.25

That's the answer I was getting as well... of course, this wouldn't be the first time I've seen wrong answers in a text book.

**Craka** · September 11th 2008, 02:56 AM

I agree with you about text books. Though this is a question from an academic institution. Would be nice if things were checked before being handed out.

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