How do you find the equation of the line tangent to #y=(x^2)(e^3x)# at the point where x=1/3?

Question

Christopher Agresta · Accepted Answer

Equation of tangent is #27y-9e^3x+2e^3=0# As #y=(x^2)(e^3x)=e^3x^3# hence at #x=1/3#, #y=e^3(1/3)^3=e^3/27# i.e. curve passes through #(1/3,e^3/27)#. Now #(dy)/(dx)=3e^3x^2# and hence as it gives the slope of the tangent at #(x,y)#, slope is #3e^3(1/3)^2# or #e^3/3# Hence equation of line passing through #(1/3,e^3/27)# and slope #e^3/3# is #(y-e^3/27)=e^3/3(x-1/3)# or #27y-e^3=9e^3x-3e^3# or #27y-9e^3x+2e^3=0#

Isaiah Allen · Answer

undefined To find the equation of the line tangent to the curve y = (x^2)(e^3x) at the point where x = 1/3, we need to find the slope of the tangent line at that point and then use the point-slope form of a line to write the equation.

1. Find the derivative of the function y = (x^2)(e^3x) using the product rule and chain rule.
2. Evaluate the derivative at x = 1/3 to find the slope of the tangent line at that point.
3. Use the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope, to write the equation of the tangent line.

Let's go through the steps:

1. Differentiate y = (x^2)(e^3x) using the product rule and chain rule:
   - The derivative of x^2 is 2x.
   - The derivative of e^3x is 3e^3x.
   - Applying the product rule, the derivative of (x^2)(e^3x) is (2x)(e^3x) + (x^2)(3e^3x).

2. Evaluate the derivative at x = 1/3:
   - Substitute x = 1/3 into the derivative: (2(1/3))(e^(3(1/3))) + ((1/3)^2)(3e^(3(1/3))).
   - Simplify the expression to find the slope of the tangent line at x = 1/3.

3. Use the point-slope form of a line to write the equation of the tangent line:
   - Substitute the values of the point (x1, y1) and the slope m into the point-slope form: y - y1 = m(x - x1).
   - Replace y1 with the value of y at x = 1/3 from the original function.
   - Replace x1 with the given x-coordinate, which is 1/3.
   - Simplify the equation to find the equation of the tangent line.

That's it! You have found the equation of the line tangent to y = (x^2)(e^3x) at the point where x = 1/3.