C4 Questions I'm stuck on...

Q1: imgur.com/Yp404yY

I am stuck on parts c and d for this question.
For part a I did:
dA/dt= dA/dR x dr/dt, da/dt=2(pi)r
Hence, da/dt= 2(pi)r.dr/dt

For part b:
Da/dt=k[(pi)a^2 - (pi)r^2]
2(pi)r.dr/dt= k(pi)a^2 -k(pi)r^2
2r.dr/dt = k(a^2 - r^2)

I can't do part c as it requires expressing r in terms of 3 letters which I don't know how to do. I tried turning it into 2 letters by introducing some auxiliary element, but that failed.

Q2 & Q3 (parts c and d for both only): I could not find any examples of these types of questions + was never covered in the textbook :frown:

imgur.com/G7lMk52
imgur.com/LqobQx9

Q4: imgur.com/5lJkYMW

Pretty much don't know where to begin at all with this question, it's at the very end of the exercise so it's probably very difficult.

Q5: imgur.com/4oygahs

So after showing the general solution is x= Ae^-kt...

For part a of this question (before t=0...) If t=1 that means Q=Qa
T=0 : Q=a (Quantity q of drug is administered)
T<0 : Q=0
x=a???

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Reply 1

ghostwalker

Original post by anon360

Q1: imgur.com/Yp404yY

I can't do part c as it requires expressing r in terms of 3 letters which I don't know how to do. I tried turning it into 2 letters by introducing some auxiliary element, but that failed.

Q1. Part d is going to depend on part c.

For c, note that k and a are both constants. The only variables are t and r.

Your D.E. is no different, in principle, to 2r.dr/dt = 5(3^2 - r^2), which I imagine you can solve.

Reply 2

anon360

Original post by ghostwalker

So I treat k and a as if I only have 1 constant?

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Reply 3

ghostwalker

Original post by anon360

So I treat k and a as if I only have 1 constant?

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Well, they're both constants and appear in different places in your equation, so you can't combine them into one.

Reply 4

anon360

Bump.

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Reply 5

RDKGames

Study Forum Helper

Original post by anon360

Bump.

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Q1 So you have

2r\frac{dr}{dt}=k(a^2-r^2)

which by seperation of variables leads you to

\frac{2r}{a^2-r^2}.dr=k .dt

Now integrate both sides giving

\displaystyle \int \frac{2r}{a^2-r^2} .dr = \int k .dt

I assume you can handle the RHS and for LHS the trick here is to turn it into

\displaystyle -\int \frac{-2r}{a^2-r^2} .dr

where now the numerator is the derivative of the bottom with respect to r. Keep in mind that

a

and

k

are constants therefore you can treat them as if they are both the number 1.

Q2 C So find the position vector of A at t=0 and t=10 then substract one from another to find the displacement. Do the same for B. Show that these vectors are perpendicular.

Q2 D A has position vector as

r_A=(1)i+(3)j+(4-\frac{1}{10}t)k

and B has position

r_B=(1+\frac{1}{5})i+(-3)j+(2)k

so find the vector AB by doing B-A and hence find the magnitude. Once you have the magnitude, you can use differentiation to minimise the distance and find the value of t at which it occurs - hence then find the actual distance.

Q3 C Find the equation of the line through A and B with vector AB and point B as the starting point to notice what happens.

Q3 D Get the direction vector of OP. Get the direction vector of AB. For them to be perpendicular, you need

\cos(\theta)=\frac{\mathbf{a} \cdot \mathbf{b}}{\lvert \mathbf{a}\lvert \lvert \mathbf{b} \lvert}=0

so the dot product must be 0 and solve for lambda from there.

Q4 The direction has to be the same therefore the gradients must be the same and since the gradient of the line is