Question

A fish is being reeled in by a person sitting on the dock. The tip of the fishing rod, 15 feet above the water, with the fish and the water surface form a perfect right triangle. The fishing line was reeling the fish at a rate of negative 4 feet per second (negative rate because the fishing line is getting shorter and shorter).

a. How fast is the distance from the fish to the dock getting shorter at the instant when there is a total of 25 feet of fishing line out?
b. At that moment as above, what rate (in radians per second) is the angle of elevation (the angle formed between the line and the water) changing?
c. Also, what is the rate change of the area of the right triangle at that moment?
Note: To get full credit you must draw the picture (-1 point if not), and label A for the vertical side/the height, and B for the horizontal side /the base, and C for the hypotenuse of the right triangle, and as the angle of elevation (-1 point if sides are labeled differently and -1 point for missing unit in the answer)

Answer 1

(a)10 ft/sec

(b)0.12 rad/sec

(c)
`-75 ft^2/sec`

Explanation:

Using Pythagoras Theorem

`C^2=A^2+B^2`

The distance A is always constant.

Taking derivatives with respect to time, we have:

`2C (dC)/(dt)= 2B (dB)/(dt)`

(a) When C=25 feet and A=15 feet

`25^2=15^2+B^2\\B^2=25^2-15^2\\B^2=400\\B=20 $ feet`

dC/dt= -4 ft/sec

`2C (dC)/(dt)= 2B (dB)/(dt)`

`2(25) (-4)= 2(20) (dB)/(dt)\\\\-400= 40 (dB)/(dt)\\\\(dB)/(dt) =-10 ft/sec`

Therefore, the rate (dB/dt) at which the distance from the fish to the dock getting shorter is: 10 ft/sec

(b)

Using Trigonometry

`\sin \theta = (A)/(C)`

`(d)/(dt) (\sin \theta) = (d)/(dt)(A)/(C)\\\\\cos \theta (d\theta)/(dt) = (C(dA)/(dt)-A(dC)/(dt))/(C^2)\\\\ (d\theta)/(dt) = (-A(dC)/(dt))/(C^2\cos \theta)`

When A = 15, C=25 feet

`\cos \theta = (20)/(25)`

Therefore:

`(d\theta)/(dt) = (-15(-4))/(25^2 * (20)/(25))\\(d\theta)/(dt)=(60)/(500)\\(d\theta)/(dt)=0.12$ rad/sec`

The angle of elevation (the angle formed between the line and the water) is changing at a rate of 12 rad/sec.

(c)

Area of the Triangle,
`X=\frac12 BA`

`(dX)/(dt) =\frac12(B(dA)/(dt)+A(dB)/(dt))\\\\(dX)/(dt) =\frac12A(dB)/(dt)`

Therefore, the rate at which the area of the right triangle is changing is:

`(dX)/(dt) =\frac12(15)(-10)\\\\(dX)/(dt)=-75 ft^2/sec`

The area of the right triangle is decreasing at a rate of 75 square feet per second.