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International Baccalaureatte | Maths SL
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Can someone help me explain how to differentiate z=v*v' with respect to t?
Date Posted: 5 years ago
Views: 569
Realise that :
d/dt (v dv/dt)
= v d²v/dt² + (dv/dt)(dv/dt)
= v (d²v/dt²) + (dv/dt)²
So,
v (d²v/dt²) + (dv/dt)² - 3t = 0
v (d²v/dt²) + (dv/dt)² = 3t
∫ (v (d²v/dt²) + (dv/dt)²) = ∫ 3t dt
v dv/dt = 3/2 t² + C, C is a constant
sub t = 0, v(0) = 1, v'(0) = 0,
C = 0
You'll have to integrate again here
∫ v dv/dt dt = ∫ 3/2 t² dt
½v² = ½t³ + D, D is a constant
Sub t = 0, v(0) = 1, v'(0) = 0,
½(1²) = ½(0³) + D
D = ½
So ½v² = ½t³ + ½
v² = t³ + 1
v = √(t³ + 1) since v > 0.
So t³ > -1 for real values of v.
This means t > -1^(⅓) = -1 for real values of v
d/dt (v dv/dt)
= v d²v/dt² + (dv/dt)(dv/dt)
= v (d²v/dt²) + (dv/dt)²
So,
v (d²v/dt²) + (dv/dt)² - 3t = 0
v (d²v/dt²) + (dv/dt)² = 3t
∫ (v (d²v/dt²) + (dv/dt)²) = ∫ 3t dt
v dv/dt = 3/2 t² + C, C is a constant
sub t = 0, v(0) = 1, v'(0) = 0,
C = 0
You'll have to integrate again here
∫ v dv/dt dt = ∫ 3/2 t² dt
½v² = ½t³ + D, D is a constant
Sub t = 0, v(0) = 1, v'(0) = 0,
½(1²) = ½(0³) + D
D = ½
So ½v² = ½t³ + ½
v² = t³ + 1
v = √(t³ + 1) since v > 0.
So t³ > -1 for real values of v.
This means t > -1^(⅓) = -1 for real values of v
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I have forgotten a fair bit of my maths knowledge already, hopefully my answer helps.
Date Posted:
5 years ago
Now, v is a function of t. In other words, v depends on t.
When we differentiate t with respect to t, we obtain 1. But when we differentiate v with respect to t, we do not obtain 1, but rather, dv/dt.
Similarly, when we differentiate dv/dt with respect to t we obtain d2v/dt2.
When we differentiate t with respect to t, we obtain 1. But when we differentiate v with respect to t, we do not obtain 1, but rather, dv/dt.
Similarly, when we differentiate dv/dt with respect to t we obtain d2v/dt2.
why is the derivative of v*v' with respect to t = v*v''+ (v')^2
im confused because when i use the chain rule, im usually multiplying the outside function with inside function
Now, v multiplied by dv/dt is an expression which cannot be simplified in general, unless you know the exact expression for v and dv/dx.
So v*dv/dt is a product of two variables involving v and dv/dt (these two terms are related just because one is the derivative of the other, otherwise it’s just the product of two unrelated terms.
Whenever we differentiate a product of two functions of x, we use the product rule (you have learnt this last year or two years ago probably).
v is the “first term”, dv/dt is the “second term”.
By the product rule, differentiating v dv/dt with respect to t gives
“Keep the first, differentiate the second” + “Keep the second, differentiate the first”
v * d2v/dt2 + (dv/dt) (dv/dt)
But both the dv/dt mean the same thing
Hence we have v * d2v/dt2 + (dv/dt)^2
So v*dv/dt is a product of two variables involving v and dv/dt (these two terms are related just because one is the derivative of the other, otherwise it’s just the product of two unrelated terms.
Whenever we differentiate a product of two functions of x, we use the product rule (you have learnt this last year or two years ago probably).
v is the “first term”, dv/dt is the “second term”.
By the product rule, differentiating v dv/dt with respect to t gives
“Keep the first, differentiate the second” + “Keep the second, differentiate the first”
v * d2v/dt2 + (dv/dt) (dv/dt)
But both the dv/dt mean the same thing
Hence we have v * d2v/dt2 + (dv/dt)^2
The chain rule here is masked under “dv/dt”, because we are unsure of what v actually is in terms of t.
When we differentiate t with respect with t, we obtain 1 as the result.
When we differentiate v with respect to t, we obtain 1 (because v looks like t) but then we do the chain rule here to obtain dv/dt). Hence the result is dv/dt.
Similarly when we differentiate v2, we obtain 2v times dv/dt.
You can verify this by letting, say, v = 3x + 1 so that v2 becomes (3x + 1)^2 and then differentiate this expression.
When we differentiate t with respect with t, we obtain 1 as the result.
When we differentiate v with respect to t, we obtain 1 (because v looks like t) but then we do the chain rule here to obtain dv/dt). Hence the result is dv/dt.
Similarly when we differentiate v2, we obtain 2v times dv/dt.
You can verify this by letting, say, v = 3x + 1 so that v2 becomes (3x + 1)^2 and then differentiate this expression.
This is probably your first time trying to differentiate terms like v dv/dt. In this case we use the product rule to differentiate this term. Differentiating v gives you dv/dt, while differentiating dv/dt gives d2v/dt2.
It does take time to get used to these expressions though. Let me know if you need more help in this and I will do my best to explain them again.
It does take time to get used to these expressions though. Let me know if you need more help in this and I will do my best to explain them again.
Yes thank you very much, i cannot solve this unless i use the product rule. What did you mean by using chain rule then? others have also told me to use chain rule to solve but i was stuck there
The chain rule here is hidden, and it’s hard to see. More appropriate is the product rule in this instance.
When we differentiate v2 with respect to t,
We obtain (2) (v)^2 * (differentiate the v).
The “differentiate the v” part is the chain rule.
When we differentiate v2 with respect to t,
We obtain (2) (v)^2 * (differentiate the v).
The “differentiate the v” part is the chain rule.
The second part in which I try to integrate v dv/dt to obtain the 0.5 v2 thing is the chain rule part, but done in reverse. It takes experience for me to observe this, and I do forget them myself from time to time.
In short, this question requires you to do two series of integrations. The first one is direct integration after substitution, while the second is the reverse of the chain rule differentiation.
In general, if a function is of the form
(y^n) (dy/dx)
and we integrate this with respect to x, we obtain y^(n + 1) whole thing divided by n.
You can try differentiating y^2, y^3, y^4, y^5 etc with respect to x. At the end of each of them, you will do the chain rule to get something times dy/dx, and this is where the chain part occurs.
In general, if a function is of the form
(y^n) (dy/dx)
and we integrate this with respect to x, we obtain y^(n + 1) whole thing divided by n.
You can try differentiating y^2, y^3, y^4, y^5 etc with respect to x. At the end of each of them, you will do the chain rule to get something times dy/dx, and this is where the chain part occurs.
v dv/dt is the derivative of 0.5 v2 with respect to t upon inspection (the extra dv/dt term is what distinguishes the derivative of x^n from the derivative of v^n). So once you realise this, we proceed to integrate both sides of the equation, and you will have v2 in terms of t. Square root both sides to get v in terms of t.
Could you take a look at at this question https://ask.manytutors.com/questions/61268
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