I would appreciate some help with the following question:
Find an equation of the tangent line to curve $y=e^{-x}$ which is perpendicular to the line $2x-y=8$
[Math] Finding equation of tangent to the curve perpendicular to the line
calculus
calculus
I would appreciate some help with the following question:
Find an equation of the tangent line to curve $y=e^{-x}$ which is perpendicular to the line $2x-y=8$
Best Answer
The line $2x-y = 8$ has a slope of $2$. Any line perpendicular to it must have slope $-\frac{1}{2}$.
The derivative of $y=e^{-x}$ evaluated at a given value of $x$ will give you the slope of the tangent line to to the graph of $y=e^{-x}$ at that value of $x$. So we wish to solve $\frac{dy}{dx} = - \frac{1}{2}$ for $x$, and we thus know that the tangent line at this value of $x$ has slope $-\frac{1}{2}$ as desired.
To finish up, we can write an equation for a line provided we know its slope and a single point through which it passes. We figured out the desired slope back in the beginning. How can we find a point that it passes through? Draw a picture of what's going on if you're stuck here.