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Night Photography - Street Light Question?

the burst is because of the aperture... set the aperture to f/8 and see what happens... it is all about practice and developing your own skills and knowledge... 1. A street light is at the top of a 18 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of?

Suppose the woman is distance 2x from the foot of the lamp post. This means the tip of the shadow is 3x away from the foot of the light. (Draw a diagram and you will see - use similar triangles). So the speed of the woman is d(2x)/dt = 2dx/dt = 8 feet/sec so that dx/dt = 4 feet/sec The speed of the shadow is d(3x)/dt = 3dx/dt = 3 x 4 = 12 feet per second. It would seem that the distance from the light is not needed - unless I have made an error. Hope not.

2. how do you find the shadow of a 5 ft tall person if he or she is 30 ft away from a 20 foot tall street light?

hmmm... if theyre both congruent triangles, you should be able to compare them, right? so i would say 20 is to 30 as 5 is to x. which comes out to 7.5. did i do something wrong? what am i missing? 3. A street light is at the top of a 11 ft tall pole. A woman 6 ft?

| |5. |___ |..x..| |....6|... |......|..... |___.|____ ...x.....s draw diagram like the one above and let x be the distance of woman from pole and let s be the length of shadow. The two little triangles are similar so their gradients are equal 5/x = 6/s so s = (6/5)x differentiate both sides with respect to time gives ds/st = (6/5)dx/dt and your given dx/dt = 5ft/s so ds/dt=6ft/s is the speed of the shadow it's constant and does not depend on x so the 35ft is useless (note: the above answer mistakenly found the velocity of woman plus shadow 56=11) 2. Here it's best to just think of one of them moving rather than both ie consider the situation from A's perspective, A is stationary at the origin of your graph and sees B at x=50mi east of him travelling east (x) at dx/dt=17mi/hr and north (y) at dy/dt=25mi/hr so the position of B relative to A, by integration if you like, as a function of time is x = 5017t y = 25t the distance between them is D = (x y) = ((5017t) (25t)) = plug in t=5 is the answer 3. given dV/dt = 20ft^3/min and r/h = constant = k say so r = kh substitute in V = (1/3)pi r^2h = (1/3)pi k^2h^3 DV/dt = pi k^2h^2dh/dt need to know the k = r/h ratio to solve this one ,.,.,.,

4. A street light is mounted at the top of a 15 foot tall pole. A man 6 ft tall walks away from the pole with a s?

We are given the height of the pole (15 ft), the height of the man (6 ft), and the distance between them in terms of t (x=6t). How fast the tip of the shadow moving is the rate of change (over time, t) in the distance from the pole to the tip. Let D(t)=distance from the pole to the tip. D(t) = x y so D(t) = 6t y Now we know x in terms of t but we still need to get y in terms of t. We can do this by getting y in terms of x. You can see by drawing the information you are given in a diagram as I have, that x and y are related. In the diagram you should see two triangles. The dimensions of the two triangles are related as follows: 15/D = 6/y and D = x y so 15/(xy) = 6/y solve for y in terms of x y=(2/3)x so we can rewrite D=xy in terms of x only D(t)=x(2/3)x =(5/3)x we are given x=6t so D(t)= (5/3)(6t) = 10t Now that we have the distance in terms of t we can find the rate of change in the distance by taking the derivative of D(t) D'(t)=10 In some cases where the rate changes over time, you would need to plug your distance (40 ft) into D(t) as D(t), solve for t, and then plug t into D'(t), however in this case D'(t) is a constant 10 ft/s.

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