A spider is descending vertically at a rate of 0.5 cm/sec. A lizard sits patiently on the ground at a spot 15 cm feet from the shadow of the spider (assume the shadow is directly below the spider). At what rate is the spider’s angle of elevation, θ, decreasing when it is 5 cm above the ground?

Question

I'm stuck, please help. Thanks in advance!

Answer 1

as usual, draw a diagram. If the spider is at height y, then

y/15 = tanθ

1/15 dy/dt = sec^2θ dθ/dt
when y=5, tanθ = 1/3, so sec^2θ = 1+tan^2θ = 10/9

Now jut plug in your numbers and solve for dθ/dt