How do you find the equation of the tangent and normal line to the curve
#y=x^3# at x=1?

Question

How do you find the equation of the tangent and normal line to the curve 
#y=x^3# at x=1?

Christopher Agresta · Accepted Answer

The equation of the tangent #y=3x-2#
Then the equation of the normal is #y=-1/3x+4/3#

Given -

#y=x^3#

At #x=1; y= 1^3=1#

#(1,1)#
It is at this point there is a tangent and a normal.
The slope of the tangent is equal to the slope of the given curve at #x=1#
The slope of the curve at any given point is, its first derivative.

#dy/dx=3x^2#

Slope of the curve at #x=1#

#m=3xx1^1=3#

Then the slope of the tangent #m_1=3#
The equation of the tangent

#y-y_1=m_1(x-x_1)#
#y-1=3(x-1)#
#y-1=3x-3#
#y=3x-3+1#
#y=3x-2#

If the two lines cut vertically then #m_1 xx m_2=-1#

#m_2=(-1)/(m_1)=(-1)/3=-1/3#

Then the equation of the normal is -

#y-y_1=m_1(x-x_1)#
#y-1=-1/3(x-1)#
#y-1=-1/3x+1/3#
#y=-1/3x+1/3+1#
#y=-1/3x+4/3#

Isabella Ackerman · Answer

undefined To find the equation of the tangent line to the curve $y = x^3$ at $x = 1$, we first find the derivative of the function $y = x^3$. The derivative is $y' = 3x^2$.

At $x = 1$, the slope of the tangent line is $y'(1) = 3(1)^2 = 3$.

Thus, the equation of the tangent line is $y - y_1 = m(x - x_1)$, where $m = 3$ and $(x_1, y_1)$ is the point $(1, 1^3) = (1, 1)$.

Substituting these values into the equation, we get $y - 1 = 3(x - 1)$, which simplifies to $y = 3x - 2$.

To find the equation of the normal line, we use the fact that 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 $-\frac{1}{3}$.

Using the point-slope form again with the point $(1, 1)$, we get $y - 1 = -\frac{1}{3}(x - 1)$, which simplifies to $y = -\frac{1}{3}x + \frac{4}{3}$.

How do you find the equation of the tangent and normal line to the curve #y=x^3# at x=1?