Can someone guide me along this calculus question?

Question

Near a buoy, the depth of a lake at the point with coordinates (x, y) is z = 200 + 0.02x^2 - 0.001y^3, where x, y, and z are measured in meters. A fisherman in a small boat starts at the point (80, 60) and move toward the buoy, which is located at (0, 0). Is the water under the boat getting deeper or shallower when he departs? Explain.

steiner1745, I think it is in the directional derivatives section too.

I found the Duf(x, y), the vector towards the buoy, then substitute the vector to get Duf(x0, y0). But since the answer is a constant, how do I show whether it's getting deeper or shallower using what I've found?

Answer 1

Hint: Look up "directional derivative" in
your calculus book and see what you find.

Answer 2

You could directly substitute the values in the equation to get the answer.

Let B = The depth at the buoy, F = Depth at fisherman

B = 200 + 0.02 (0)^2 - 0.001(0)^3
B = 200 + 0 - 0
B = 200

F = 200 + 0.02(80)^2 - 0.001(60)^3
F = 200 + 0.02( 6400) - 0.001(216000)
F = 200 + 128 - 216
F = 112

Hence, the water is getting deeper.

Answer 3

it easy not a calc problem just plug and chug z=200+.o2x^2-.001y^3
plug in 80 for x .02(80)^2 and 60 for y .001(60)^3 and u get that as x gets bigger u get higher and as y gets bigger u get lower at a slower rate
so it starts out geting higher but as the cubed catches on with squared and starts subtracting form 200meters

Answer 4

Wow Ling, your avatar shows you to appear to be so young. You should be lucky to be so interested, and so gifted enough mentally, to have the wherewithal to be able to tackle such complicated math questions. You should be proud, and I wish you the best of luck in whatever you choose to do in life.

Answer 5

NO!!!

Answer 6

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