Calculus Word Problem-Related Rates

Question

A shark, looking for dinner, is swimming parallel to a straight beach and 40 feet offshore. The shark is swimming at the constant speed of 6 feet per second. At time t=0, the shark is directly opposite a lifeguard station. How fast is the shark moving away from the lifeguard station when the distance between them is 50 feet?

Answer 1

Here is a diagram to help you visualize as I explain: https://i.imgur.com/gLGo4Ao.png

 

At t=0, the shark is 40 ft directly away from the lifeguard tower. At another time t, we can say that he is x ft away from where he started.

 

Using the Pythagorean theorem, we have that his distance function from the tower is s(t)=√[x²+40²]=√x²+1600). (I'm using the letter s instead of d, so we don't confuse it with the d in derivatives.) However, it will make it easier to differentiate if we square both sides giving us: s²=x²+1600. Notice that both x and s are changing as t changes. Since they are both functions of t, we will need to use the chain rule when we implicitly differentiate them.

 

Now we implicitly differentiate:

d/dt[s²]=d/dt[x²+1600]

2s*(ds/dt)=2x*(dx/dt)

s*ds/dt=x*(dx/dt)

 

The question is telling us that we care about the speed away from the lifeguard tower, ds/dt, at the time when it's 50 feet away as it moves at a speed of 6 ft/s. So what are the values of s and dx/dt? Can you figure out what the value of x must be at this time?

 

I'll let you do it from there. Let me know if you have any questions.