If x=3t^-2 and y= e^-10t^2 compute dy/dx

[ugu]

Help correct error please

 

[BigBeachBanana]

Try dy/dt again, remember to use chain rules.

 

[blamocur]

Note 1 : You got the sign wrong when computing the derivative for [imath]y(t)[/imath], but the sign l
Note 2 : You can modify the fraction in the result so that [imath]t^{-3}[/imath] is no longer in the denominator. Th

Try dy/dt again, remember to use chain rules.

Good catch, and sorry I've missed that error.

 

[ugu]

Try dy/dt again, remember to use chain rules.

can you just show me the input so as not to be confused

 

[blamocur]

can you just show me the input so as not to be confused

What is meant is that [imath]\frac{d}{dt} e^{-10t^2}[/imath] is not equal to [imath]-10e^{-10t^2}[/imath]. If you don't see this right away I suggest you write out the chain rule explicitly for that expression.

 

[BigBeachBanana]

can you just show me the input so as not to be confused

Hope you can finish the rest
[math]\frac{d}{dt}(\mathrm{e}^{-10t^2}) = \mathrm{e}^{-10t^2}* \frac{d}{dt}(-10t^2)=?[/math]

 

[ugu]

Hope you can finish the rest
[math]\frac{d}{dt}(\mathrm{e}^{-10t^2}) = \mathrm{e}^{-10t^2}* \frac{d}{dt}(-10t^2)=?[/math]

Is -10t^2 not suppose to cancel the other -10t^2 in the bracket at the right?

 

[BigBeachBanana]

Is -10t^2 not suppose to cancel the other -10t^2 in the bracket at the right?

In general:
[math]\frac{d(\mathrm{e}^{u})}{dt}=\mathrm{e}^{u}*\frac{du}{dt}[/math]Apply it a simpler problems:
[math] \frac{d}{dt}(\mathrm{e}^{2t}) = \mathrm{e}^{2t}* \frac{d(2t)}{dt}= \mathrm{e}^{2t}*2[/math]Apply it to your problem

 

[ugu]

d/dt(e^2t) =e^-10t^2 * du/dt

 

[BigBeachBanana]

I think you meant:

d/dt(e^(-10t^2)) =e^-10t^2 * du/dt

[math]\frac{d}{dt}(\mathrm{e}^{-10t^2}) = \mathrm{e}^{-10t^2}* \frac{d}{dt}(-10t^2)[/math]The last thing you'll have to do is [imath]\frac{d}{dt}(-10t^2)[/imath], what is this equal to?

 

[ugu]

d/dt(-10t^2) = 10t^1/2 ? where did we put the x=3t^-2 in the question?

 

[ugu]

and how much is private tutorial at all as it is certain i will fail the course come January exam. My tutor teaches this very fast and like English

 

[BigBeachBanana]

d/dt(-10t^2) = 10t^1/2 ? where did we put the x=3t^-2 in the question?

This is incorrect, this is a simple power rule differentiation.
What's [imath]\frac{d(\mathrm{x}^{2})}{dx} =?[/imath]

 

[ugu]

This is incorrect, this is a simple power rule differentiation.
What's [imath]\frac{d(\mathrm{x}^{2})}{dx} =?[/imath]

= x

 

[BigBeachBanana]

= x

Incorrect. You used the power rule correctly when you did [imath]\frac{d(3t^{-2})}{dt}[/imath]. It's the same idea.

 

[ugu]

Mean d2x/dt. ?

 

[BigBeachBanana]

Use the same thought process you used to find [imath]\frac{d}{dt}(3t^{-2})[/imath], to find [imath]\frac{d}{dt}(-10t^2)[/imath]

 

[pka]

Given [imath]x=3t^{-2}~\&~y=e^{-10t^2}[/imath] compute [imath]\dfrac{dy}{dx}[/imath]
[imath]\dfrac{dx}{dt}=-6t^{-3}~\&~\dfrac{dy}{dt}=(-20t)e^{-10t^2}[/imath]
[imath]{\large \dfrac{dy}{dx}=\dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}}=\dfrac{(-20t)e^{-10t^2}}{-6t^{-3}}=\dfrac{10}{3}t^4e^{-10t^2}[/imath]
[imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath]

 

[ugu]

Given [imath]x=3t^{-2}~\&~y=e^{-10t^2}[/imath] compute [imath]\dfrac{dy}{dx}[/imath]
[imath]\dfrac{dx}{dt}=-6t^{-3}~\&~\dfrac{dy}{dt}=(-20t)e^{-10t^2}[/imath]
[imath]{\large \dfrac{dy}{dx}=\dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}}=\dfrac{(-20t)e^{-10t^2}}{-6t^{-3}}=\dfrac{10}{3}t^4e^{-10t^2}[/imath]
[imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath]

excellent sir. But i need to get this. It means in the second line with is the first step. we multiplied 3 by -2 and the power has to reduce by 1- which became -3. Also for the dy/dt, we multiply -10 by +2 by why didn't its power increase nor reduced?

 

[BigBeachBanana]

I think you're confused between the derivative of polynomials and exponentials. Let's look at simpler problems [imath]t^2[/imath] vs. [imath]e^{2t}[/imath].
[imath]t^2[/imath] is a polynomial where your variable t is at the base, whereas [imath]e^{2t}[/imath] is an exponential function where your t variable is in the exponent. Does this make sense?