How do you find an equation of the tangent and normal line to the curve #y=2xe^x# at the point (0,0)?

Question

Ezekiel Alexander · Accepted Answer

tangent line: #y = 2x#

normal line: #y = -1/2 x#

#y = 2xe^x, y(0) = 0#

For the tangent line, we will use the derivative:

Using the product rule:
#y' = 2e^x + 2x e^x = 2e^x (1+x), y'(0) = 2#

The tangent line is therefore #y = 2x#

The normal to line #y = mx + c# has slope #-1/m#. The normal also passes through the origin so it's equation is #y = -1/2 x#

Elijah Abram · Answer

undefined To find the equation of the tangent and normal line to the curve y=2xe^x at the point (0,0), we need to find the slope of the tangent line and the slope of the normal line at that point.

First, we find the derivative of the curve y=2xe^x using the product rule and chain rule. The derivative is given by dy/dx = 2e^x + 2xe^x.

Next, we substitute x=0 into the derivative to find the slope of the tangent line at the point (0,0). Substituting x=0 into dy/dx gives us the slope of the tangent line, which is 2e^0 = 2.

The slope of the normal line is the negative reciprocal of the slope of the tangent line. Therefore, the slope of the normal line is -1/2.

Now that we have the slopes, we can use the point-slope form of a line to find the equations of the tangent and normal lines.

For the tangent line, we use the point-slope form with the point (0,0) and the slope 2. The equation of the tangent line is y - 0 = 2(x - 0), which simplifies to y = 2x.

For the normal line, we use the point-slope form with the point (0,0) and the slope -1/2. The equation of the normal line is y - 0 = (-1/2)(x - 0), which simplifies to y = -1/2x.

Therefore, the equation of the tangent line is y = 2x and the equation of the normal line is y = -1/2x.