Question Number 203638 by thean0000 last updated on 24/Jan/24
Commented by thean0000 last updated on 24/Jan/24
$${help}\:{me} \\ $$$$ \\ $$
Commented by thean0000 last updated on 24/Jan/24
$${which}\:{one}\:{is}\:{answer} \\ $$
Answered by esmaeil last updated on 24/Jan/24
![z=x^2 y+3xy^4 { ((x=sin2t)),((y=cost)) :}→(dz/dt)]_(t=0) =? z=4sin^2 tcos^3 t+6sintcos^5 t→(dz/dt)]_(t=0) = 4(2sintcos^4 t−3cos^2 tsin^3 t)+ 6(cos^5 t−5cos^4 tsin^2 t)= (0+0+0)+6(1−0)=6](https://www.tinkutara.com/question/Q203641.png)
$${z}={x}^{\mathrm{2}} {y}+\mathrm{3}{xy}^{\mathrm{4}} \\ $$$$\left.\begin{cases}{{x}={sin}\mathrm{2}{t}}\\{{y}={cost}}\end{cases}\rightarrow\frac{{dz}}{{dt}}\right]_{{t}=\mathrm{0}} =? \\ $$$$\left.{z}=\mathrm{4}{si}\overset{\mathrm{2}} {{n}tcos}^{\mathrm{3}} {t}+\mathrm{6}{sintcos}^{\mathrm{5}} {t}\rightarrow\frac{{dz}}{{dt}}\right]_{{t}=\mathrm{0}} = \\ $$$$\mathrm{4}\left(\mathrm{2}{sintcos}^{\mathrm{4}} {t}−\mathrm{3}{cos}^{\mathrm{2}} {tsin}^{\mathrm{3}} {t}\right)+ \\ $$$$\mathrm{6}\left({cos}^{\mathrm{5}} {t}−\mathrm{5}{cos}^{\mathrm{4}} {tsin}^{\mathrm{2}} {t}\right)= \\ $$$$\left(\mathrm{0}+\mathrm{0}+\mathrm{0}\right)+\mathrm{6}\left(\mathrm{1}−\mathrm{0}\right)=\mathrm{6} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by MM42 last updated on 24/Jan/24
$$\frac{{dz}}{{dt}}\:=\frac{{dz}}{{dx}}×\frac{{dx}}{{dt}}+\frac{{dz}}{{dy}}×\frac{{dy}}{{dt}} \\ $$$$=\left(\mathrm{2}{xy}+\mathrm{3}{y}^{\mathrm{4}} \right)×\mathrm{2}{cos}\mathrm{2}{t}+\left({x}^{\mathrm{2}} +\mathrm{6}{xy}\right)\left(−{sint}\right) \\ $$$${t}=\mathrm{0}\Rightarrow\frac{{dz}}{{dt}}=\mathrm{6}\:\:\checkmark \\ $$$$ \\ $$