Question

use parametric differentiation \[x=t ^{3}\times \sin(t)\] and \[y=e ^{2t} - \ln \left| 5t \right|\]
tried this multiple times can't seem to arrive at the proposed answer

Answer 1

Remember:
\[\frac{ dy }{ dx }=\frac{ \frac{ dy }{ dt } }{ \frac{ dx }{ d } }\]

Answer 2

I seem to always come to:

\[\frac{ 2e ^{2t} -\frac{ 1 }{ x }}{ 3t ^{2} \times \cos(t)-t ^{3}\sin(t) }\]

Answer 3

yes I sure do remember that @FutureMathProfessor

Answer 4

On the bottom you should be adding the terms and on the top you should have 1/t

Answer 5

however the proposed answer is:
\[\frac{ dy }{ dx }\] =\[\frac{ 2te ^{2t}-1 }{ t ^{3}(3\cos(t)-tsin(t)) }\]

I don't understand how the \[t ^{3}\] appears two times in the answer?
@FutureMathProfessor

Answer 6

@oldrin.bataku could you please help me out with this? :S

Answer 7

dy/dt= (2e^2t) - (1/5t).. right?

Answer 8

yes that is correct but where do you get \[\frac{ 1 }{ 5t }\] from?

Answer 9

y=e^2t−ln|5t|.. thats the equation right? differentiation of ln|x| is 1/x.. so diff of ln|5t| is 1/5t..

Answer 10

I put that in the calculator and it differentiates to \[\frac{ dy }{ dt } = \frac{ 1 }{ t }\]

Answer 11

eh, my bad. sorry. it is 1/t. log(5t)= log5 + logt.. differentiation gives you 1/t.

Answer 12

\[x=t^3\sin(t) \\ \frac{dx}{dt}=t^3\cos(t)+\sin(t)(3t)^2\]
\[y=e^{3t}-\ln(5t) \\ \frac{dy}{dt}=3e^{3t}-\frac{1}{\cancel5t}(\cancel5)\]
Chain rule
\[\frac{dy}{dt}=\frac{dy}{dx} \cdot \frac{dx}{dt}\]
\[\frac{dy}{dx}=\frac{dy}{dt} \div \frac{dx}{dt}\]
\[\frac{dy}{dx}=\frac{3e^{3t-\frac{1}{t}}}{t^3\sin(t)} \\ \frac{dy}{dx}=\frac{3te^{3t}-1}{t^4\sin(t)}\]