@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]
) 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.