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A light is at the top of a pole $50$ft high. A ball is dropped from the same height from a point $30$ft away from the light. How fast is the shadow of the ball moving along the ground $1\over 2$ seconds later? (Assume the ball falls a distance $s = 16t^2$ft in $t$ seconds.)

The way I attempted at this is as so:

The geometrical picture we are looking at is of two heights ($= 50$ft) $30$ft horizontally away from each other. One of the heights decreases towards the horizontal at a rate $-16t^2$. As this height decreases, connect the tips of both heights with a line and produce this line till it hits the horizontal. Call the length between the point at which the line hit the horizontal and the foot of the decreasing height $x$. Then the tip of this length is the shadow of the ball, and it is required to find the rate at which this length decreases. The image below should provide a visual aid:

enter image description here

Call the balls height from the horizontal $y$ (this is $50 - 16t^2$ on the picture), then by what is given, ${dy\over dt}=-32t$ (since $y = 16t^2$)

An obvious relation between the variables and the constants is of the proportionality between the sides of the triangles.

${x\over y} = {x + 30 \over 50}$

Solving for $x$, $x = {30y \over 50 - y}$

Implicitly differentiating, $x^\prime = 30y(50 - y)^{-2}y^\prime + (50 - y)^{-1}30y^\prime$

Substituting for $y^\prime = -32$, we get $x^\prime = -1500$.

Is this result correct? I must ask for I do not consider myself yet experienced, hence I have to get constant affirmation from those more professional. Thank you in advance.

Camelot823
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    this is a fine answer – C Squared Apr 17 '23 at 15:09
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    There is a small error in your calculation. You should have instead $dy/dt = -32t$. Substitute this into your expression for $dx/dt$ and everything is fine. – M. Wind Apr 17 '23 at 15:55
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    Also consider whether $30y(50 – y)^{–2}(-32) + (50 – y)^{-1}30(-32)$ is actually $-1500.$ (It isn't, so either you made an additional error after substituting $-32$ for $y',$ or you actually substituted $y'=-32t$ but forget the $t$ in two places in your work. – David K Apr 17 '23 at 16:05
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    \over is a tricky macro, as you discovered. I prefer \frac, for example, \frac{x}{y}. Also, whatever you were using for the negative sign, just use an ordinary old ASCII hyphen instead; it will typeset much better. – David K Apr 17 '23 at 16:13
  • @M.Wind ahh, that must have been a typo, it didn't disrupt the actual calculation I made. Thank you for noticing! – Camelot823 Apr 17 '23 at 16:21
  • @DavidK yes, in my actual calculation I made the correct substitution, and then substituted $t = {1\over 2}$. I forgot to include it in detail in the post. – Camelot823 Apr 17 '23 at 16:25
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    I thought it was a bit too much of a coincidence to have two mistakes cancel out, especially since I couldn't guess what the second mistake was. – David K Apr 17 '23 at 16:48

1 Answers1

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That's fine, as an alternative we have

$$\frac x{50-16t^2}=\frac{x+30}{50} \implies x(t)=\frac{30(50-16t^2)}{16t^2}=\frac{375}{4t^2}-30$$

therefore

$$x'(t)= -2 \frac{375}{4t^3} \implies x'(1/2)=-1500 \;ft/s$$

user
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