Question #271922

Find the equation of Tangent line to the curve ;


x=e^t , y=e^-t at t=1

Expert's answer

Solution;

To find the equation of the tangent we need the slope;

m=dydxm=\frac{dy}{dx} At the point of tangency (xo,yo)(x_o,y_o)

Then the equation of tangent is ;

(y−yo)=m(x−xo)(y-y_o)=m(x-x_o)

So we compute;

dydt=−e−t\frac{dy}{dt}=-e^{-t}

dxdt=et\frac{dx}{dt}=e^t

Therefore;

dydx=dydtdtdx=−e−tet=−e−2t\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=\frac{-e^{-t}}{e^t}=-e^{-2t}

At t=1;

m=−0.135m=-0.135

x0=e1=2.718x_0=e^{1}=2.718

yo=0.368y_o=0.368

Hence, equation of tangent is;

(y−0.368)=−0.135(x−2.718)(y-0.368)=-0.135(x-2.718)

y=3.086−0.135xy=3.086-0.135x





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