FRAC{{d\left( {{e^X}} \right)}}{{dx}} = {e^x}\)\(\frac{{d\left( {\sin x} \right)}}{{dx}} = \cos x\)\(\frac{{d\left( {\cos x} \right)}}{{dx}} = \; - \sin x\)Parametric Differentiation:If x = f(t), y = g(t), where t is a parameter, then \(\frac{{dx}}{{dy}} = \frac{{f'\left( t \right)}}{{g'\left( t \right)}}\)CALCULATION:Given: x = et cost and y = et sintFirst let's find out dx/dt⇒ \(\frac{dx}{dt} =e^t cos t -e^t sin t\)Now let's find out dy/dt⇒ \(\frac{dy}{dt} = e^t sint + e^t cos t\)⇒ \(\frac{dx}{dy} =\frac{e^t cos t \ - \ e^t sin t}{e^t sint \ + \ e^t cos t}\)⇒ \({\left[ {\frac{{dy}}{{dx}}} \right]_{t= 0}} = \frac{1 - 0}{0+1} = 1\)Hence, correct OPTION is 2.

"> FRAC{{d\left( {{e^X}} \right)}}{{dx}} = {e^x}\)\(\frac{{d\left( {\sin x} \right)}}{{dx}} = \cos x\)\(\frac{{d\left( {\cos x} \right)}}{{dx}} = \; - \sin x\)Parametric Differentiation:If x = f(t), y = g(t), where t is a parameter, then \(\frac{{dx}}{{dy}} = \frac{{f'\left( t \right)}}{{g'\left( t \right)}}\)CALCULATION:Given: x = et cost and y = et sintFirst let's find out dx/dt⇒ \(\frac{dx}{dt} =e^t cos t -e^t sin t\)Now let's find out dy/dt⇒ \(\frac{dy}{dt} = e^t sint + e^t cos t\)⇒ \(\frac{dx}{dy} =\frac{e^t cos t \ - \ e^t sin t}{e^t sint \ + \ e^t cos t}\)⇒ \({\left[ {\frac{{dy}}{{dx}}} \right]_{t= 0}} = \frac{1 - 0}{0+1} = 1\)Hence, correct OPTION is 2.

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If x = et cost and y = et sint, then what is \(\dfrac{dx}{dy}\) at t = 0 equal to?

UPSEE Calculus in UPSEE 10 months ago

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CONCEPT:\(\FRAC{{d\left( {{e^X}} \right)}}{{dx}} = {e^x}\)\(\frac{{d\left( {\sin x} \right)}}{{dx}} = \cos x\)\(\frac{{d\left( {\cos x} \right)}}{{dx}} = \; - \sin x\)Parametric Differentiation:If x = f(t), y = g(t), where t is a parameter, then \(\frac{{dx}}{{dy}} = \frac{{f'\left( t \right)}}{{g'\left( t \right)}}\)CALCULATION:Given: x = et cost and y = et sintFirst let's find out dx/dt⇒ \(\frac{dx}{dt} =e^t cos t -e^t sin t\)Now let's find out dy/dt⇒ \(\frac{dy}{dt} = e^t sint + e^t cos t\)⇒ \(\frac{dx}{dy} =\frac{e^t cos t \ - \ e^t sin t}{e^t sint \ + \ e^t cos t}\)⇒ \({\left[ {\frac{{dy}}{{dx}}} \right]_{t= 0}} = \frac{1 - 0}{0+1} = 1\)Hence, correct OPTION is 2.

Posted on 10 Dec 2024, this text provides information on UPSEE related to Calculus in UPSEE. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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