Answered: "How do you find the equation of the..."

Quinn Dean

Answered question

2022-04-14

How do you find the equation of the line tangent to

y = (x^{2}) (e^{3} x)

at the point where x=1/3?

cadhail6n1t

Beginner2022-04-15Added 14 answers

Explanation:
As

y = (x^{2}) (e^{3} x) = e^{3} x^{3}

hence at

x = \frac{1}{3}, y = e^{3} {(\frac{1}{3})}^{3} = \frac{e^{3}}{27}

i.e. curve passes through

(\frac{1}{3}, \frac{e^{3}}{27})

.
Now

\frac{d y}{d x} = 3 e^{3} x^{2}

and hence as it gives the slope of the tangent at (x,y),
slope is

3 e^{3} {(\frac{1}{3})}^{2}

\frac{e^{3}}{3}

Hence equation of line passing through

(\frac{1}{3}, \frac{e^{3}}{27})

and slope

\frac{e^{3}}{3}

(y - \frac{e^{3}}{27}) = \frac{e^{3}}{3} (x - \frac{1}{3})

27 y - e^{3} = 9 e^{3} x - 3 e^{3}

27 y - 9 e^{3} x + 2 e^{3} = 0

Do you have a similar question?

Recalculate according to your conditions!