allthebest Posted September 3, 2016 Report Share Posted September 3, 2016 Help with this calculus problem please A 5 foot girl is walking toward a 20 foot lamp post at a constant rate of 6 ft/sec. The light from the lamp post causes the girl to cast a shadow. how fast is the tip of her shadow moving? Quite confused with the fact that the girl is moving towards, so is it minus? I would love it if there would be an explanation! Thank you Reply Link to post Share on other sites More sharing options...
haese225 Posted September 3, 2016 Report Share Posted September 3, 2016 The hard part is understanding that there are two variables. Here, I denoted k to be the distance from the pole to the girl, and from the question dk/dt (speed of girl) is 6 ft/sec. I denoted x to be the distance from pole to shadow tip. I believe the question wants us to find dx/dt (shadow tip speed). See attached: 2 Reply Link to post Share on other sites More sharing options...
allthebest Posted September 4, 2016 Author Report Share Posted September 4, 2016 8 hours ago, 4lan said: The hard part is understanding that there are two variables. Here, I denoted k to be the distance from the pole to the girl, and from the question dk/dt (speed of girl) is 6 ft/sec. I denoted x to be the distance from pole to shadow tip. I believe the question wants us to find dx/dt (shadow tip speed). See attached: Thank you for your explanation! But I'm still quite confused.. isn't is supposed to be negative since its going towards the lamp...? Or shouldn't the x-k part of your picture be 6 ft/sec?? I'm so confused Reply Link to post Share on other sites More sharing options...
haese225 Posted September 4, 2016 Report Share Posted September 4, 2016 1 hour ago, allthebest said: Thank you for your explanation! But I'm still quite confused.. isn't is supposed to be negative since its going towards the lamp...? Or shouldn't the x-k part of your picture be 6 ft/sec?? I'm so confused I guess you have two questions? 1. The dk/dt part is 6 ft/s because the question states that "the girl walks to the pole". This requires a rate of change of distance of the girl to the pole (dk/dt) 2. Even though k is decreasing, doesn't mean dk/dt should be negative. Note that the sign (+ or -) depends on the reference. For example, in my diagram the distance k is measured from the girl to the pole, or say, to from left to right. Thus, anything moving to the right (-->) is positive. Sorry, I think I need to change my x and k (as to avoid confusion). Let, x = distance from shadow tip -> pole (instead of pole -> shadow tip) k = distance from girl -> pole I think the takeaway is that the sign (+ or -) depends on the reference. 2 Reply Link to post Share on other sites More sharing options...
allthebest Posted September 4, 2016 Author Report Share Posted September 4, 2016 35 minutes ago, 4lan said: I guess you have two questions? 1. The dk/dt part is 6 ft/s because the question states that "the girl walks to the pole". This requires a rate of change of distance of the girl to the pole (dk/dt) 2. Even though k is decreasing, doesn't mean dk/dt should be negative. Note that the sign (+ or -) depends on the reference. For example, in my diagram the distance k is measured from the girl to the pole, or say, to from left to right. Thus, anything moving to the right (-->) is positive. Sorry, I think I need to change my x and k (as to avoid confusion). Let, x = distance from shadow tip -> pole (instead of pole -> shadow tip) k = distance from girl -> pole I think the takeaway is that the sign (+ or -) depends on the reference. Ohhh I get it! Thank you for your explanations!! 1 Reply Link to post Share on other sites More sharing options...
Vioh Posted September 4, 2016 Report Share Posted September 4, 2016 Referring to @allthebest post on this thread below, the answer from the book is actually 2 ft//sec. However, this is because the wording of the question is a bit ambiguous, not because @4lan did something wrong. In fact, he got everything correct. It's just that the question wants us to find the speed of the shadow with respect to the person walking. In other words, we have to find d(x-k)/dt. d(x – k)/dt = dx/dt – dk/dt = 8 – 6 = 2 ft/sec 3 Reply Link to post Share on other sites More sharing options...
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