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Fish On!

In Ripple on a Calm Lake we looked at the rate of change of the area of a ripple created when we cast a fishing lure into the water. As a continuation of that problem, we will now look at what happens when next if we manage to get a fish on the line. This will give us more help with related rates with different equations. Note that we are going to use the same process as outlined in Ripple in a Calm Lake, so I would suggest looking over the process given there before following along with this problem.

How fast do you have to reel?

Suppose that you are standing on the shore of a calm lake on a beautiful spring day. You cast a fishing lure out into the lake and it causes a ripple. You begin to real the line it trying to entice a fish into biting the lure. Suddenly, you feel a tug on your line as a fish hits the lure. You set the hook and begin reeling in. As you reel in, the fish decides to run and starts swimming directly toward you at a rate of 15 feet per second. In order to keep the fish on the line, you need to keep tension on the line. Therefore, you have to reel in order to keep the line taught. Currently there is 100 feet of line let out past your pole and the tip of the pole is 10 foot above the fishes height. At what rate do you have to reel line in so that you keep the line taught?

Read problem

As we read through the problem a few times, we want to pay particular attention to anything that will be helpful. While the first five sentences help to set a scene, we don’t get to anything that will help us with the mathematical model until the sixth sentence. We then get the following

  • The fish is swimming toward you at 15 feet per second.
  • There is currently 100 feet of line out.
  • The tip of the pole is 10 feet above the height of the fish.
  • We want to know the rate at which we have to reel in line.

As a final step in the drawing, we should draw a line connecting the fish and the fisherman. We then have the following picture.

Variables

We now have to find all of our variables. As we do this, we want to think about what is changing in this problem. Here, we note that the length of the fishing line is changing, we will therefore assign \(l=\)line length. The distance from the fish to the point below the pole will also be changing as the fish swims. We will assign this as \(f=\)fish’s distance. The other thing that is changing is time, so we let \(t=\)time.

Note that the distance from the tip of the pole to the point below it will not change. Therefore, instead of assigning this a variable, we will just call it 10. It is also helpful to note that, currently, \(l=100\). It may be helpful to find other variables or information, but we will do this as needed later on.

Rates

Now that we know our variables, we want to find write out the rates that we know.

  • We know that the distance from the fish to the point below the pole is changing at 15 feet per second.
  • That is, \(\frac{df}{dt}= \pm 15\).
  • Since the fish is swimming toward us, \(f\) is getting smaller.
  • Therefore, \(\frac{df}{dt}=-15\).
  • We want to know the rate of change of \(l\) with respect to time.
  • That is, we need to find \(\frac{dl}{dt}\).

Equation

We now have to find an equation relating the variables in our problem. Since we have the length of two sides of a right triangle, the Pythagorean theorem would be a good choice for an equation. We then have that
\begin{align*}
f^{2}+10^{2}=l^{2}.
\end{align*}

Even though we know that \(l\) is currently 100, we have to leave this as variable in the equation because it is changing. We can, however, use 10 for the second leg because this is not changing.

Implicitly Differentiate

Now that we have an equation, we can implicitly differentiate. Recall that we want to find \(\frac{dl}{dt}\). Therefore, we will take the derivative of both sides of our equation with respect to \(t\). This will give us
\begin{align*}
\frac{d}{dt}(f^{2}+10^{2})=\frac{d}{dt}(l^{2}).
\end{align*}

Now, note that since \(f\) and \(l\) are functions of time, we will have to use the chain rule to find that \(\frac{d}{dt}f^{2}=2f\frac{df}{dt}\) and \(\frac{d}{dt}l^{2}=2l\frac{dl}{dt}\). Since \(10\) is constant, \(10^{2}\) is constant, so \(\frac{d}{dt}10^{2}=0\). We now get
\begin{align*}
\frac{d}{dt}(f^{2}+10^{2})&=\frac{d}{dt}(l^{2}) \\
2f\frac{df}{dt}&=2l \frac{dl}{dt}.
\end{align*}

Substitute

Note that by implicitly differentiating, we have found an equation involving \(\frac{df}{dt}\) and \(\frac{dl}{dt}\). This was the goal of using this model. Now that we have done this, we can now substitute in what we know and solve for what we want. We then have, at the current point in time,
\begin{align*}
2f\frac{df}{dt}&=2l \frac{dl}{dt}. \\
2 f (-15)&=2 (100) \frac{dl}{dt}.
\end{align*}

Here, we note that we want to find \(\frac{dl}{dt}\), so we shouldn’t have something to plug in here. However, we are not trying to find \(f\), so we do need something to plug in. Because we need \(f\) to finish our problem we will need to find that using the information we are given. In particular, we know that \(f^{2}+10^{2}=l^{2}\). Noting that \(l\) is currently 100, we get that
\begin{align*}
f^{2}+10^{2}&=100^{2} \\
f^{2}&=10,000-100 \\
f^{2}&=9900 \\
f& \approx 99.5.
\end{align*}

Using this value for \(f\), we now get that
\begin{align*}
2(99.5)(-15)&=2(100)\frac{dl}{dt} \\
\frac{dl}{dt}& \approx -14.9.
\end{align*}

Solution

Here we note that we have found the solution to the model we created of the problem, but we haven’t given the solution to the original question. Therefore, we have to convert this back into the original question. In order to do this, we have to determine what is going on.

  • We have \(\frac{dl}{dt} \approx -14.9\)
  • Since line length is in inches and time is in seconds, \(\frac{dl}{dt}\) is in inches per second.
  • Because our answer is negative, the length of the line is decreasing.

If we combine all of this together, we find that the length of the line is decreasing at a rate of \(14.9\) feet per second. In a more colloquial way, we would say, “The length of the fishing line is getting smaller by 14.9 feet per second.” If we take this a step further to give it in terms of the original question we find that, “As the fish is swimming towards us at 15 feet per second, we will have to reel the line in at 14.9 feet per second in order to keep the line taught.”

Conclusion

I hope seeing this helped you get more comfortable with relates rates. If it did, make sure to like the post and share on Social Media. You can also find more Calculus help here.

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