calculus related rates word problem | Mathematics | Page 1

proven pagoda Apr 8, 2024, 9:41 PM

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this is the answer key but i dont understand how to like get to the conclusion or show my work and stuff

drifting egretBOT Apr 8, 2024, 9:41 PM

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proven pagoda this is the answer key but i dont understand how to like get to the conclusion o...

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muted hatch Apr 8, 2024, 9:43 PM

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proven pagoda this is the answer key but i dont understand how to like get to the conclusion o...

Try drawing a picture. Also, let the height of the person and the lamppost be h and H, and the speed of person walking from the pole be v. You need to find the rate u of how quickly the shadow grows when the person is at distance x from the pole.

proven pagoda Apr 8, 2024, 9:43 PM

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okay

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like this?

muted hatch Apr 8, 2024, 9:47 PM

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Rather, like this.

proven pagoda Apr 8, 2024, 9:47 PM

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the main issue i am having is i dont know where she got (x+y)/20 from too and i dont know how to label the components

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ohhhh okay

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so i am solving for v then?

muted hatch Apr 8, 2024, 9:47 PM

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proven pagoda the main issue i am having is i dont know where she got (x+y)/20 from too and i ...

Better not to use actual values.

muted hatch Apr 8, 2024, 9:48 PM

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proven pagoda so i am solving for v then?

No, you know v. You need to find u = d(Δx)/dt.

proven pagoda Apr 8, 2024, 9:48 PM

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ohhh right my bad because v=16

muted hatch Apr 8, 2024, 9:48 PM

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No.

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v = 5 ft/s.

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16 m is the value of x.

proven pagoda Apr 8, 2024, 9:49 PM

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oh okay because 16 is the length h is away from H

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okay i get that i think

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so trianglex is the length of the shadow correct

muted hatch Apr 8, 2024, 9:52 PM

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What?

proven pagoda Apr 8, 2024, 9:52 PM

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like sorry

proven pagoda Apr 8, 2024, 9:52 PM

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muted hatch Rather, like this.

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this is the shadow right

muted hatch Apr 8, 2024, 9:54 PM

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Oh. That's delta x.

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And yes, that's the length of the shadow.

proven pagoda Apr 8, 2024, 9:54 PM

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oh my bad!

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okay thanks

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so im not sure how to solve the problem still what is the equation i have to take the derivative of

muted hatch Apr 8, 2024, 9:56 PM

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proven pagoda so im not sure how to solve the problem still what is the equation i have to tak...

First of all, note that the small and big triangles are similar. So, try writing the ratio of their corresponding sides.

proven pagoda Apr 8, 2024, 9:57 PM

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im so sorry i dont really know what you mean!

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like dh/dH?

muted hatch Apr 8, 2024, 9:58 PM

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No, don't worry about differentiating for now.

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Green and red triangles are similar.

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Thus, their ratios of base to height are the same.

proven pagoda Apr 8, 2024, 10:02 PM

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im sorry i still dont get it like what im supposed ot do

muted hatch Apr 8, 2024, 10:03 PM

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What I mean is that due to similarity of triangles we have:
Δx/h = (x + Δx)/H

proven pagoda Apr 8, 2024, 10:03 PM

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OH

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okay yes i get that

muted hatch Apr 8, 2024, 10:04 PM

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Alright.
Now, multiply both parts by H and express Δx.

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So, what did you get?

proven pagoda Apr 8, 2024, 10:07 PM

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deltax= hx/H-h

muted hatch Apr 8, 2024, 10:08 PM

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If you meant Δx = hx/(H - h), then yes.

proven pagoda Apr 8, 2024, 10:08 PM

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yes i did!

muted hatch Apr 8, 2024, 10:08 PM

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Nice!
Now, differentiate both parts with respect to t. Note that h and H are constants.

proven pagoda Apr 8, 2024, 10:08 PM

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so since theyre constants that means their derivatives are 0 right?

muted hatch Apr 8, 2024, 10:09 PM

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Well, yes, but what I meant, rather, is that you can treat the right side as (h/(H - h))x and recall that if a is a constant, then (ay)' = ay'.

proven pagoda Apr 8, 2024, 10:09 PM

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okay!

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wait so should i plug in their values

muted hatch Apr 8, 2024, 10:11 PM

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Not yet.

proven pagoda Apr 8, 2024, 10:11 PM

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okay so then should i just use quotient rule to find derivative

muted hatch Apr 8, 2024, 10:12 PM

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No.

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There are no quotients of functions here.

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You just have Δx = kx, where k = h/(H - h) is a constant.

proven pagoda Apr 8, 2024, 10:13 PM

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okay

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so but then how would i find k' without quotient rule

muted hatch Apr 8, 2024, 10:14 PM

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Again, k is a constant, so (d/dt)(kx) = k(dx/dt).

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Remember: derivative is linear.

proven pagoda Apr 8, 2024, 10:14 PM

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okay so what should my next step be

muted hatch Apr 8, 2024, 10:15 PM

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Well, as Δx = hx/(H - h), that means d(Δx)/dt = h(dx/dt)/(H - h).

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But dx/dt is just v.

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So, d(Δx)/dt = hv/(H - h).

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And that's what we need.

proven pagoda Apr 8, 2024, 10:16 PM

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v=5ft right

muted hatch Apr 8, 2024, 10:16 PM

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v = 5 ft/s, rather.

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Don't forget to use the correct units.

proven pagoda Apr 8, 2024, 10:16 PM

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okay so should i write ds/dt=(h(5ft/s))/(H-h)

muted hatch Apr 8, 2024, 10:17 PM

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Well, we denoted the length of the shadow by Δx.
Also, why substitute just v? We've already got the formula, you can substitute everything.

proven pagoda Apr 8, 2024, 10:17 PM

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OH okay nows the time to plug in stuff then right

muted hatch Apr 8, 2024, 10:18 PM

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Yeah. As usual the algorithm of solving pretty much any physics and chemistry problem is:

Derive the general formula.
Substitute the given values.

proven pagoda Apr 8, 2024, 10:19 PM

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OMG i got the answer

muted hatch Apr 8, 2024, 10:19 PM

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Nice!

proven pagoda Apr 8, 2024, 10:19 PM

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thank you so so much

muted hatch Apr 8, 2024, 10:19 PM

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You're welcome!

proven pagoda Apr 8, 2024, 10:20 PM

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but i also have another similar problem do i submit another thread for that?

muted hatch Apr 8, 2024, 10:20 PM

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Doesn't matter, so do whatever you prefer.

proven pagoda Apr 8, 2024, 10:20 PM

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okay ill send it here

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i sent it in the server yesterday but i didnt really understand and i kind of wanna start it over

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ive got like the same issue where idk how to label things and where to go next

muted hatch Apr 8, 2024, 10:28 PM

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Ah, I see.
First, let's also generalize this. We have the following picture. We know x, y and dy/dt, and we want to find dl/dt.

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First of all, let's write the equation connecting x, y and l.

proven pagoda Apr 8, 2024, 10:29 PM

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okay

muted hatch Apr 8, 2024, 10:29 PM

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Well, I'm sure you know what that equation will be, considering that this is a right triangle.

proven pagoda Apr 8, 2024, 10:30 PM

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so y= 2400, x=700

muted hatch Apr 8, 2024, 10:30 PM

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No, no.

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As usual, we forget about values.

proven pagoda Apr 8, 2024, 10:30 PM

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yes y^2+x^2=l^2

muted hatch Apr 8, 2024, 10:30 PM

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Those will be important only in the end.

proven pagoda Apr 8, 2024, 10:30 PM

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alrightt

muted hatch Apr 8, 2024, 10:30 PM

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Yes.

proven pagoda Apr 8, 2024, 10:30 PM

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okay

muted hatch Apr 8, 2024, 10:30 PM

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We have x^2 + y^2 = l^2. Try differentiating both sides with respect to t.

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Note that x is constant.

proven pagoda Apr 8, 2024, 10:31 PM

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how would i know x is constant like if i was given this problem on my own

muted hatch Apr 8, 2024, 10:31 PM

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Well, the person doesn't move.

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So, x is constant.

proven pagoda Apr 8, 2024, 10:31 PM

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okayy

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so when i diffrentiate do i do dx/dt to every value

muted hatch Apr 8, 2024, 10:32 PM

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You take d/dt of both sides.

proven pagoda Apr 8, 2024, 10:32 PM

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oh ok no second letter in the numerator then

muted hatch Apr 8, 2024, 10:33 PM

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Not sure what you mean. We are just differentiating both sides with respect to t.

proven pagoda Apr 8, 2024, 10:33 PM

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okay no its fine i get what ur saying

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so then 2y+2x=2l

muted hatch Apr 8, 2024, 10:34 PM

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No. Again, x = const, so (d/dt) x^2 = 0.

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And you forgot to use the chain rule.

proven pagoda Apr 8, 2024, 10:34 PM

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oh yes my bad

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wait how come i have to use chain rule

muted hatch Apr 8, 2024, 10:35 PM

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Because both y and l are functions of t.

proven pagoda Apr 8, 2024, 10:35 PM

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so i have to use chain rule with d/dt and y^2

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and also d/dt l^2 ?

muted hatch Apr 8, 2024, 10:35 PM

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Yes.

proven pagoda Apr 8, 2024, 10:35 PM

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okay one sec

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can i use product rule

muted hatch Apr 8, 2024, 10:37 PM

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Not sure where you would use it here. You just need the chain rule.

proven pagoda Apr 8, 2024, 10:37 PM

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ohh ur right ok

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so then d/dt 2y= d/dt 2l

muted hatch Apr 8, 2024, 10:38 PM

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Well, you forgot to write y and l there. So, rather 2y(dy/dt) = 2l(dl/dt).

proven pagoda Apr 8, 2024, 10:38 PM

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okay got it

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so now what am i supposed to do

muted hatch Apr 8, 2024, 10:39 PM

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Now, express dl/dt from here.

proven pagoda Apr 8, 2024, 10:40 PM

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(dy/dt 2y)/2l?

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= dl/dt

muted hatch Apr 8, 2024, 10:41 PM

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Well, you can cancel the twos. So, dl/dt = (y/l)(dy/dt).

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Now, the only thing we don't know yet is l.

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For that, we remember that l^2 = x^2 + y^2. So, what is l?

proven pagoda Apr 8, 2024, 10:42 PM

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2500

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i think

muted hatch Apr 8, 2024, 10:43 PM

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No values.

proven pagoda Apr 8, 2024, 10:43 PM

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ohhh

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so l=sqrt(x^2+y^2)

muted hatch Apr 8, 2024, 10:44 PM

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Right.

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So, dl/dt = (y/√(x^2 + y^2))(dy/dt).

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And we know everything here, so you can find dl/dt.

proven pagoda Apr 8, 2024, 10:44 PM

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okayy

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then u plug in at the end ?

muted hatch Apr 8, 2024, 10:44 PM

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Yes.

proven pagoda Apr 8, 2024, 10:44 PM

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kk one second

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dl/dt=(sqrt(y^2+x^2))/(y+x)

muted hatch Apr 8, 2024, 10:48 PM

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Wait, where did that come from?

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We already got dl/dt = (y/√(x^2 + y^2))(dy/dt). You just need to substitute the given values.

proven pagoda Apr 8, 2024, 10:49 PM

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OH

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what would i substitute for dy/dt tho

muted hatch Apr 8, 2024, 10:50 PM

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We are given dy/dt = 900 ft/s.

proven pagoda Apr 8, 2024, 10:51 PM

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answer i got 864

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omg thats correct

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thank you so much sorry i was such a hassle

#calculus related rates word problem