Getting the Best View

Calculus Level 3

A 6 foot tall painting is placed on the wall so that the very bottom of the painting sits 7 feet above the floor. Suppose your eyes are 5 feet above the floor and that you are standing 6 feet from the painting. Let T be your line of sight with the top of the painting and B be your line of sight with the bottom of the painting. What is the absolute value of the rate the angle formed by T and B is changing (in radians per second) as you approach the painting at 3 feet per second?

The answer is 0.09.

1 solution

Andrew Ellinor
Oct 5, 2015

Let $x$ be your horizontal distance to the wall. The angle in question is $\alpha = \text{arccot}\left(\frac{x}{8}\right) - \text{arccot}\left(\frac{x}{2}\right)$

Differentiating both sides with respect to time yields $\frac{d\alpha}{dt} = \left(-\frac{8}{x^2 + 64} + \frac{2}{x^2 + 4}\right)\frac{dx}{dt}$

Now we plug in $x = 6$ and $\frac{dx}{dt} = 3$ into $\frac{d\alpha}{dt}$ and find that it's equal to $-0.09$ .

I too found 0.09 rad/s, Speed is -3 m/s.

Kris Hauchecorne - 4 years, 9 months ago

Hi dx/dt = -3ft/sec as we approch the painting and X is decreasing. thus the correct answer should be +0.09 rad/sec

dror lev - 5 years ago

i think it should be "the man is 6feet from the wall"

Terrell Bombb - 4 years, 11 months ago

Since x decreases dx/dt = -3 m/s... this gives d@/dt as Positive 0.09 which names sense since the angle increases as you approach the painting

geoff taylor - 3 years, 6 months ago

Getting the Best View

The answer is 0.09.

1 solution

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