Question Help (1 Viewer)

Jojofelyx · Aug 12, 2021

Need help rearranging this with respect to the rates of change.

Etho_x · Aug 12, 2021

Etho_x · Aug 12, 2021

Etho_x said:
View attachment 31577

Although usually I wouldn't have done it this way from the get go. I'd write out the chain rule for the rate of change in interest first and then through that try to figure out what you can do with the information provided.

Jojofelyx · Aug 12, 2021

Etho_x said:
Although usually I wouldn't have done it this way from the get go. I'd write out the chain rule for the rate of change in interest first and then through that try to figure out what you can do with the information provided.

Thank you !!

Jojofelyx · Aug 12, 2021

2nd question here. Any help would be very appreciated.

Life'sHard · Aug 12, 2021

i/ just multiply all the roots = -c/a

ii/ u do P(alpha)=0 = sub x as alpha
since you know what alpha is from i/ sub it into the fat equation that eqauls to 0. Simplify and solve.

CM_Tutor · Aug 16, 2021

On the first question, there is absolutely nothing wrong with @Etho_x's solution, but a modified approach to consider is to start from the derivative that you seek and then expand it using the Chain Rule. For example, here we seek the rate of change of volume with time and we are given the rate of change of surface area with time:

$\cfrac{dV}{dt} = \cfrac{dV}{d(\text{something})} \times \cfrac{d(\text{something})}{dt}$

and the only "something" we have information on is surface area, so we need

$\cfrac{dV}{dt} = \cfrac{dV}{dS} \times \cfrac{dS}{dt}$

We know that $\cfrac{dS}{dt} = 24\ \text{cm}^2\ \text{s}^{-1}$ but don't have an equation linking $V$ and $S$ ... but we can link them both to the radius of the sphere, $r$ , and so can further separate

$\cfrac{dV}{dt} = \cfrac{dV}{dS} \times \cfrac{dS}{dt} = \left(\cfrac{dV}{dr} \times \cfrac{dr}{dS}\right) \times \cfrac{dS}{dt}$

$\text{Now,} \quad V = \cfrac{4}{3}\pi r^3 \quad \implies \quad \cfrac{dV}{dr} = \cfrac{4}{3} \times 3\pi r^2 = 4\pi r^2$

$\text{And,} \quad S = 4\pi r^2 \quad \implies \quad \cfrac{dS}{dr} = 4\pi \times 2r = 8\pi r \quad \implies \quad \cfrac{dr}{dS} = \cfrac{1}{8\pi r}$

$\cfrac{dV}{dt} = \cfrac{dV}{dr} \times \cfrac{dr}{dS} \times \cfrac{dS}{dt} = 4\pi r^2 \times \cfrac{1}{8\pi r} \times 24 = \cfrac{24r}{2} = 12r$

So, when the diameter is 24 cm, and thus the radius is 12 cm, the rate of change of the volume with respect to time is $12 \times 12 = 144$ cm³ s^-1.

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If you are wondering why I got $\cfrac{dV}{dt} = 12r$ while Etho_x's working show $\cfrac{dV}{dt} = r^2$ at the end, it is because my working as found $\cfrac{dV}{dt}$ at any time whereas Etho_x has used the value of $\cfrac{dr}{dt}$ specifically when $r = 12$ cm, and thus is only valid at that specific size of balloon. That's why we get the same answer for the specific case but different answers for any other case. Being ultra-picky, Eltho_x should really have substituted the value of $r$ one step earlier, at the same point as using the specific value of $\cfrac{dr}{dt}$ .

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