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Rates of change help

Yatayyat

How would I answer a question like this?

Screenshot 2020-02-04 at 20.57.54.jpg

This is what I have tried but I feel I have done part b) incorrectly which is needed for part c)

Thanks!

Reply 1

ghostwalker

Original post by Yatayyat

How would I answer a question like this?

Screenshot 2020-02-04 at 20.57.54.jpg

This is what I have tried but I feel I have done part b) incorrectly which is needed for part c)

Thanks!

For part b you've worked out the partial derivative of y wrt t assuming x is constant.

You want the total derivative since x is a function of t.

\displaystyle \frac{dy}{dt}= \frac{\partial y }{\partial t}+\frac{\partial y}{\partial x}\frac{dx}{dt}

Reply 2

Pangol

Original post by ghostwalker

For part b you've worked out the partial derivative of y wrt t assuming x is constant.

You want the total derivative since x is a function of t.

\displaystyle \frac{dy}{dt}= \frac{\partial y }{\partial t}+\frac{\partial y}{\partial x}\frac{dx}{dt}

Is an understanding of partial derivatives required? Can you not just use a combination of the product rule and the chain rule?

Reply 3

ghostwalker

Original post by Pangol

Is an understanding of partial derivatives required? Can you not just use a combination of the product rule and the chain rule?

My knowledge of mutlivariate calculus isn't too strong so I wouldn't like to make a definitive statement, though in this case, yes, it amounts to the same thing.

Reply 4

Pangol

Original post by ghostwalker

My knowledge of mutlivariate calculus isn't too strong so I wouldn't like to make a definitive statement, though in this case, yes, it amounts to the same thing.

Mine too! Far too many years since I've done anything like that. But I thought I was able to do this question using nothing beyond further maths A level (well, regular A level, but it's pushing at the very top end of that), so I wanted to get a second opinion.

Reply 5

Yatayyat

Original post by ghostwalker

For part b you've worked out the partial derivative of y wrt t assuming x is constant.

You want the total derivative since x is a function of t.

\displaystyle \frac{dy}{dt}= \frac{\partial y }{\partial t}+\frac{\partial y}{\partial x}\frac{dx}{dt}

How have you arrived to this equation? I’m sort of confused on this part

Reply 6

ghostwalker

Original post by Yatayyat

How have you arrived to this equation? I’m sort of confused on this part

I wouldn't like to try and derive it as I'm rather rusty on this.

Just think of x as being a function of t as well and just the standard rules for differentation.

Reply 7

Pangol

Original post by Yatayyat

How have you arrived to this equation? I’m sort of confused on this part

Put that aside for the moment - let's think about how you would differentiate this using a standard approach.

You have to differentiate

\sin(t)\sqrt{\ln x}

. This is a product, so using the product rule,

\frac{dy}{dt}=\sqrt{\ln x}\frac{d}{dt}(\sin(t))+\sin(t) \frac{d}{dt}(\sqrt{\ln x})

The only part of this that should be any sort of problem is the very last part,

\frac{d}{dt}(\sqrt{\ln x})

. But by using the chain rule, this becomes

\frac{d}{dx}(\sqrt{\ln x})\frac{dx}{dt}

- and you aleady know what

\frac{dx}{dt}

is.

Reply 8

RDKGames

Study Forum Helper

Original post by Yatayyat

How have you arrived to this equation? I’m sort of confused on this part

Basically, you cannot 'threat sqrt ln(x) as a constant' in the process of differentiation with respect to t.

This is because x is a non-constant function of t (the hint is that dx/dt is non zero!)

So, you need to use implicit differentiation along with product rule to differentiate y with respect to t.

(edited 4 years ago)

Reply 9

Yatayyat

Thanks so I have obtained dy/dt now from using implicit differentiation which had involved using product rule and chain rule when sqrt(lnx) wrt x

I used my dy/dt expression in part c) to get a k value of 2*sqrt(ln2)?

Is this okay?

(edited 4 years ago)

Reply 10

Yatayyat

Here is another Q that I have tried also

Screenshot 2020-02-05 at 19.38.38.png

This one seems to have a bit less differentiation but I have given this as my answers?

Reply 11

Pangol

Original post by Yatayyat

Thanks so I have obtained dy/dt now from using implicit differentiation which had involved using product rule and chain rule when sqrt(lnx) wrt x

I used my dy/dt expression in part c) to get a k value of 2*sqrt(ln2)?

Is this okay?

That's what I got, so I hope it's right!

If I were you, I would use the expression for dx/dt from (a) at the end of (b). Your dy/dt isn't really finished otherwise. There's no need to wait until you're working out the value of k in (c).