Find the equation to the tangent line to the curve y=3x^2-x^3 at the point (1,2).

step by step solution -:

The given equation of the curve is:

y=3x^{2}-x^{3}

Let assume that y=f(x), it follows that:

f(x)=3x^{2}-x^{3}

Find the derivative:

\frac{d}{dx}{f(x)}=f'(x)=6x-3x^2

since it is given that the tangent line to the given curve is at point (1,2) so substitute (x,y)=(1,2) into the above equation to find the slope of the tangent, it follows

f'(\textcolor{primary}{1})=6(\textcolor{primary}{1})-3(\textcolor{primary}{1})^{2}\implies f'(1)=3

So the tangent line to the curve has the slope 3 at the point (1,2)

by point-slope form, we know that equation of the line is

(y-y_1)=m(x-x_1)

where,

m = slope of the tangent

 (\textcolor{secondary}{x_1},\textcolor{tertiary}{y_1})=(\textcolor{secondary}{1},\textcolor{tertiary}{2}),m=3

on substituting the above expression we get

y-\textcolor{tertiary}{2}=3\cdot(x-\textcolor{secondary}{1})

after simplifying the above expression we get

3x-y=1

hence, the required equation of the tangent is: 3x-y=1

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