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?
I'm stuck, please help. Thanks in advance!
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
To find the rate at which the spider's angle of elevation is decreasing, we can use trigonometry. Let's first draw a diagram to visualize the situation.
spider
/|
/ |
/ | θ
15cm / |
/ |
/ | 5cm
lizard ground
Let's assign some variables:
- h: height of the spider above the ground
- x: horizontal distance from the lizard to the spider's shadow
- θ: angle of elevation between the lizard, spider, and ground
- t: time
We are given the following rates:
- dh/dt = -0.5 cm/sec (the spider's vertical descent rate)
We need to find the rate dθ/dt (the rate at which the angle of elevation is decreasing) when h = 5 cm.
Now, we can form a right triangle involving the lizard, spider, and ground. Using trigonometry, we have:
tan(θ) = (5 cm) / (15 cm + x)
To relate x and h, we can use similar triangles:
x / (15 cm) = (h - 5 cm) / 5 cm
Rearranging this equation, we can express x in terms of h:
x = (15 cm)(h - 5 cm) / 5 cm
Now, we differentiate both sides of the equation to find dx/dt:
dx/dt = (15 cm)(dh/dt) / 5 cm
Substituting dh/dt = -0.5 cm/sec, we find:
dx/dt = -1.5 cm/sec
Next, we differentiate the equation tan(θ) = (5 cm) / (15 cm + x) implicitly with respect to t:
sec^2(θ) * dθ/dt = (1 / (15 cm + x)) * dx/dt
Substituting the known values, we get:
sec^2(θ) * dθ/dt = (1 / (15 cm + (15 cm)(h - 5 cm) / 5 cm)) * (-1.5 cm/sec)
Now, we need to determine the value of sec^2(θ) when h = 5 cm. In the right triangle, the adjacent side to θ is x and the hypotenuse is 15 cm + x. So, we have:
cos^2(θ) = (15 cm) / (15 cm + x)
Substituting x = (15 cm)(h - 5 cm) / 5 cm, we can express cos^2(θ) in terms of h:
cos^2(θ) = (15 cm) / (15 cm + (15 cm)(h - 5 cm) / 5 cm)
Finally, we substitute cos^2(θ) into the equation and substitute h = 5 cm:
(15 cm / (15 cm + (15 cm)(5 cm - 5 cm) / 5 cm)) * dθ/dt = (1 / (15 cm + (15 cm)(5 cm - 5 cm) / 5 cm)) * (-1.5 cm/sec)
After simplifying and substituting the values, we can solve for dθ/dt:
dθ/dt = (-1.5 cm/sec) * ((15 cm + (15 cm)(5 cm - 5 cm) / 5 cm) / (15 cm))
By evaluating this expression, we can determine the rate at which the spider's angle of elevation is decreasing.