Related rate word problem
- Thread starter Joana
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Deleted member 4993
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Use:View attachment 27399Hello) please help me with this question (this is a related rate word problem). I can’t find the connection between t and d to solve for t so I’m totally stuck.
Thanks in advance
x^2 + y^2 = d^2 ........................... differentiate and calculate \(\displaystyle \frac{d(d)}{dt}\)
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so I tried the derivatives on both sides which turned into: dd’=2xx’ + 2yy’. The x’, x and y’ are given but not the y, and I don’t know how to find y, thus d
And I don’t know how t = 0 with help me find y. So far I only think of the Pythagorean theorem
Write down an expression (in terms of t) for x, the distance from the shore after t hours. (Using the initial distance and the speed of the boat).
Write down an expression (in terms of t) for y, the vertical height of the helicopter above the ground after t hours. (Using the speed of the helicopter).
Then the square of the distance, [MATH]d^2[/MATH], will just be [MATH]x^2 + y^2[/MATH]
Using the expressions (in terms of t) which you wrote down for x and y, write down an expression (in terms of t) for the square of the distance: [MATH]x^2 + y^2[/MATH]
You can minimise this by differentiating the expression for the square of the distance (with respect to t) and setting that equal to 0.
The time you get from this (linear) equation, is the time (in hours) when the distance is minimum.
To find the distance, just substitute that value of t back into the expression for the distance squared, and then square root that answer, to get the distance (in km).
Write down an expression (in terms of t) for y, the vertical height of the helicopter above the ground after t hours. (Using the speed of the helicopter).
Then the square of the distance, [MATH]d^2[/MATH], will just be [MATH]x^2 + y^2[/MATH]
Using the expressions (in terms of t) which you wrote down for x and y, write down an expression (in terms of t) for the square of the distance: [MATH]x^2 + y^2[/MATH]
You can minimise this by differentiating the expression for the square of the distance (with respect to t) and setting that equal to 0.
The time you get from this (linear) equation, is the time (in hours) when the distance is minimum.
To find the distance, just substitute that value of t back into the expression for the distance squared, and then square root that answer, to get the distance (in km).
I don't follow the logic.
What do you believe is the error in my method?
There is no reason to believe the minimum distance would happen at an angle of 45º. (In fact it happens at an angle less than 45º).
Why would you assert that?
I'm afraid it's not logical. The length of the hypotenuse depends on the sum of the squares of the other 2 sides, not on the sizes of the other 2 sides relative to each other.
E.g. when the two sides are the same, then they are both 100/7, giving a hypotenuse of 20.2 to 1d.p.
When the horizontal is 16 and the vertical is 12, the hypotenuse is 20.
(You seem to be assuming that the sum of the two lengths is constant, which it clearly is not).
D
Deleted member 4993
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dx/dt = - 30 and dy/dt = 40View attachment 27399Hello
Thanks in advance
x = 25 - 30*t
y = 40*t ← y = 0 at t=0
d2 = (25-30*t)2 + 1600 *t2 = 2500*t^2 - 1500*t + 625
d = √(2500*t^2 - 1500*t + 625) → d(d)/dt = 0 @ t = 3/10 hrs = 18 min → d(0.3) = dmin= 20 km
d(21m26s) = 20.20361 km ← not minimum
D
Deleted member 4993
Guest
That would be true if |dy/dt| and |dx/dt| were equal.
@Joana
I'm sorry this thread became very confusing.
Going back to my post #6
Write down an expression (in terms of t) for x, the distance from the shore after t hours. (Using the initial distance and the speed of the boat).
[MATH]\boldsymbol{x=25-30t}[/MATH]Write down an expression (in terms of t) for y, the vertical height of the helicopter above the ground after t hours. (Using the speed of the helicopter).
[MATH]\boldsymbol{y=40t}[/MATH]Then the square of the distance, [MATH]\boldsymbol{D^2}[/MATH], will just be [MATH]\boldsymbol{x^2 + y^2}[/MATH]
Using the expressions (in terms of t) which you wrote down for x and y, write down an expression (in terms of t) for the square of the distance: [MATH]\boldsymbol{D^2=x^2 + y^2}[/MATH][MATH]\boldsymbol{D^2=(25-30t)^2+(40t)^2}[/MATH]You can minimise this by differentiating the expression for the square of the distance (with respect to t) and setting that equal to 0.
[MATH]\boldsymbol{\frac{d}{dt} D^2 = 2(-30)(25-30t)+2(40)(40t)=0\\ -1500+1800t+3200t=0\\ t=0.3}[/MATH]
The time you get from this (linear) equation, is the time (in hours) when the distance is minimum.
To find the distance, just substitute that value of t back into the expression for the distance squared, and then square root that answer, to get the distance (in km).
[MATH]\boldsymbol{D^2=(25-30\times0.3)^2+(40\times0.3)^2\\ D^2=400\\ D=20}[/MATH]
Minimum distance is when t=0.3 hours (i.e. 18 mins) and minimum distance = 20km (know it is min, because 2nd derivative of distance squared is positive).
(b)
At t=0.3 x=16, y=12
[MATH]\tan \theta = \frac{y}{x}[/MATH]At t=0 [MATH]\tan\theta =\frac{3}{4}[/MATH]
[MATH]x=25-30t\\ y=40t\\ \frac{dx}{dt}=-30\\ \frac{dy}{dt}=40\\ \text{ }\\ \tan \theta = \frac{y}{x}\\ \therefore \sec^2 \theta \frac{d\theta}{dt}=\frac{x\frac{dy}{dt}-y\frac{dx}{dt}}{x^2}\\ (1+\tan^2 \theta)\frac{d\theta}{dt}=\frac{x\frac{dy}{dt}-y\frac{dx}{dt}}{x^2}\\[/MATH]
At t=0.3
[MATH]\left(1+\left(\tfrac{3}{4}\right)^2 \right) \frac{d\theta}{dt}=\frac{16(40)-12(-30)}{16^2}\\ \frac{25}{16}\frac{d \theta}{dt}=\frac{125}{32}\\ \frac{d \theta}{dt}=\frac{5}{2} \text{ (rad/h)}[/MATH]
I'm sorry this thread became very confusing.
Going back to my post #6
Write down an expression (in terms of t) for x, the distance from the shore after t hours. (Using the initial distance and the speed of the boat).
[MATH]\boldsymbol{x=25-30t}[/MATH]Write down an expression (in terms of t) for y, the vertical height of the helicopter above the ground after t hours. (Using the speed of the helicopter).
[MATH]\boldsymbol{y=40t}[/MATH]Then the square of the distance, [MATH]\boldsymbol{D^2}[/MATH], will just be [MATH]\boldsymbol{x^2 + y^2}[/MATH]
Using the expressions (in terms of t) which you wrote down for x and y, write down an expression (in terms of t) for the square of the distance: [MATH]\boldsymbol{D^2=x^2 + y^2}[/MATH][MATH]\boldsymbol{D^2=(25-30t)^2+(40t)^2}[/MATH]You can minimise this by differentiating the expression for the square of the distance (with respect to t) and setting that equal to 0.
[MATH]\boldsymbol{\frac{d}{dt} D^2 = 2(-30)(25-30t)+2(40)(40t)=0\\ -1500+1800t+3200t=0\\ t=0.3}[/MATH]
The time you get from this (linear) equation, is the time (in hours) when the distance is minimum.
To find the distance, just substitute that value of t back into the expression for the distance squared, and then square root that answer, to get the distance (in km).
[MATH]\boldsymbol{D^2=(25-30\times0.3)^2+(40\times0.3)^2\\ D^2=400\\ D=20}[/MATH]
Minimum distance is when t=0.3 hours (i.e. 18 mins) and minimum distance = 20km (know it is min, because 2nd derivative of distance squared is positive).
(b)
At t=0.3 x=16, y=12
[MATH]\tan \theta = \frac{y}{x}[/MATH]At t=0 [MATH]\tan\theta =\frac{3}{4}[/MATH]
[MATH]x=25-30t\\ y=40t\\ \frac{dx}{dt}=-30\\ \frac{dy}{dt}=40\\ \text{ }\\ \tan \theta = \frac{y}{x}\\ \therefore \sec^2 \theta \frac{d\theta}{dt}=\frac{x\frac{dy}{dt}-y\frac{dx}{dt}}{x^2}\\ (1+\tan^2 \theta)\frac{d\theta}{dt}=\frac{x\frac{dy}{dt}-y\frac{dx}{dt}}{x^2}\\[/MATH]
At t=0.3
[MATH]\left(1+\left(\tfrac{3}{4}\right)^2 \right) \frac{d\theta}{dt}=\frac{16(40)-12(-30)}{16^2}\\ \frac{25}{16}\frac{d \theta}{dt}=\frac{125}{32}\\ \frac{d \theta}{dt}=\frac{5}{2} \text{ (rad/h)}[/MATH]
Brilliant lex. AS ALWAYS!
This extra information [MATH]\frac{d\theta}{dt}[/MATH] is very important in many problems.
It did not come in my mind, we can also extract the rate of change of angle.
Thanks for showing all the work above. I hope that the OP now got the answer (idea) he/she needs, and can solve similar problems easily.
This extra information [MATH]\frac{d\theta}{dt}[/MATH] is very important in many problems.
It did not come in my mind, we can also extract the rate of change of angle.
Thanks for showing all the work above. I hope that the OP now got the answer (idea) he/she needs, and can solve similar problems easily.