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Dzidzornu Yesulagbe
:
The parametric equations of a curve are x=In(2t+3), y=(4t+2)/(2t+3). Find the gradient of the curve at the point where it crosses the y-axis
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November 6, 2016 at 3:36pm
Little Sky
:
[deqn]e^x=2t+3\\t=\frac{e^x-3}{2}[/deqn]Substitute this into y, we get[deqn]y=\frac{2e^x-4}{e^x-1}[/deqn]Differentiate using the quotient rule, we get[deqn]\frac{dy}{dx}=\frac{(e^x-1)(2e^x)-(2e^x-4)(e^x)}{(e^x-1)^2}[/deqn]After expanding the numerator,[deqn]\frac{dy}{dx}=\frac{2e^x}{(e^x-1)^2}[/deqn]By taking the limit as x approaches 0, the gradient is [deqn]\infty[/deqn]
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November 9, 2016 at 2:34am
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