Find equation of a line perpendicular to the tangent of curve at a given point.

Question

I need to find the equation to the line perpendicular to the tangent to the curve $y = x^3 -3x +1$, at the point $(2,3)$.

Our teacher assigned us homework on stuff we haven't learned, so please if you get highly technical don't be afraid to explain yourself. Thanks!

Answer 1

We know $$\frac{dy}{dx}=3x^2-3$$ so the derivate or the tangent line's slope at $(2,3)$ is $3(2)^2-3=9$, and we know that the slope of the normal is then $-1/9$. Now we have the slope and we know that on this normal, the point $(2,3)$ lies. You can now carry it out yourself.

Answer 2

Hints:

1) Calculate $\frac{\mathrm{d}y}{\mathrm{d}x}.$

2) Calulate the gradient of the tangent, $m_{T},$ by calculating $\frac{\mathrm{d}y}{\mathrm{d}x}$ at $x=2$

3) Use the fact that the gradient of the normal (the line perpendicular to the tangent line) at a point, $m_N$, is minus the reciprocal of the gradient of the tangent, at that point.

i.e. $$m_{N}=-\frac{1}{m_T}.$$

Use the equation of a straight line with gradient $m_N$ that goes through $(2,3)$, as follows:

$$y-3=m_N(x-2)$$

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Remember that $m_N$ is some number that is to be determined.

Answer 3

If $y=x^3-3x+1$ then $\frac{dy}{dx}=3x^2-3$. So $\frac{dy}{dx}$ at $(2,3)$ is $9$.

The equation of the line perpendicular and tangent to the curve is $y-3=\frac{-1}{9}(x-2)$ which when simplified is $9y+x-29=0.$

Answer 4

First, we need to find the slope of the line tangent to the curve at the point $(2,3)$.

• To do this, we need to find $\frac {dy}{dx}$ given $y = x^3 - 3x + 1$. $$\frac{dy}{dx} = 3x^2 - 3 = 3(x^2-1)$$
• So the slope of the tangent line is $\frac{dy}{dx}$ at $x = 2$: $\quad m = 3(2^2-1) = 9$.

Next, given that the line tangent to the curve at $(2, 3)$ has slope $m = 9$, we know that the slope of the line perpendicular to the tangent is the negative reciprocal of $9$:

• $m_\bot = -\dfrac 19$.

Finally, recall that given the slope m of a line, and a point $(x_0, y_0)$ on the line, we can use the point-slope form of the equation of a line:

• $y-y_0 = m(x-x_0)$
• Given that the desired perpendicular line has slope $m_\bot= - \dfrac19$, and the fact that $(2, 3)$ is a point on that line, the desired equation is given by $$y - 3 = -\frac 19(x -2)$$

Answer 5

Given a function $f(x)=x^3-3x+1$ the gradient function would be $$dy/dx$$ ie $3x^2-3$ and at (2,3) the gradient function would be equal to 9 hence for a straight line $y=mx+c$ which is tangent to the curve at (2,3) $y=9x-15$ Since the perpendicular has a gradient which is the negative reciprocal of the tangent the equation would be $$y=-x/9 + (3+2/9)$$