Question

How do you find the equation of the tangent line to the curve $y = 3{x^2} - {x^3}$ at point $(1,2)$?

Answer 1

In this equation, we have to find the tangent line to the curve $y = 3{x^2} - {x^3}$, and this can be done by performing derivatives on the variables and constants in order to solve it further, since we have both the $x$ and the $y$ variable and a slope is involved.

Complete step-by-step solution:
Here we have an equation $y = 3{x^2} - {x^3}$,
This is a curve and to find a more specific equation, we will differentiate the curve such as,
$\Rightarrow \dfrac{{dy}}{{dx}} = 6x - 3{x^2}$
Now the above is done with the formula which is:
$a \times u'(x) + b \times v'(x)$
So, substituting our values in the formula,
$3 \times \dfrac{d}{{dx}}[{x^2}] - \dfrac{d}{{dx}}[{x^3}]$
Now simplifying this further, we will break the equation down,
$\Rightarrow 3 \times 2x - 3{x^2}$
Putting into brackets,
$\Rightarrow - 3(x - 2)x$
Therefore, the equation will be,
$\Rightarrow - 3x(x - 2)$
So, the slope of tangent at $(1,2)$ is,
$\Rightarrow m = \dfrac{{dy}}{{dx}}{|_{(1,2)}} = - 3(x(x - 2))$
Now putting values of the variables that are present,
$\Rightarrow m = \dfrac{{dy}}{{dx}}{|_{(1,2)}} = - 3(1)(1 - 2)$
Simplifying the equation,
$\Rightarrow m = - 3( - 1)$
Further calculation,
$\Rightarrow m = 3$ Since the sign is removed because of the rule.
Hence, the equation of the tangent that passes through $(1,2)$ is given by,
$y = mx + c$
Substituting $y = 2$and $x = 1$, $m = 3$ in the given equation,
$\Rightarrow 2 = 3(1) + c$
Shifting constants on one side,
$\Rightarrow c = 2 - 3$
Therefore, the final value is,
$\Rightarrow c = - 1$
The equation of the tangent will be,
$\Rightarrow y = mx + c$
Putting in value of $m$ and $c$,
$\Rightarrow y = 3x + ( - 1)$
Taking out brackets,
$\Rightarrow y = 3x - 1$
Shifting variables on one side and constants on other,
$\Rightarrow 3x - y = 1$

Therefore, for an equation $y = 3{x^2} - {x^3}$ , the equation of tangent at curve will be $3x - y = 1$.

Note: The tangent line to the curve at a given point is the line which intersects the curve at the point and has the same instantaneous slope as the curve at the point. And, when we want to find the tangent line, we will need to find the slope of the curve at the point we are interested in.