14) Find the equations of the tangent and normal to the given curves at the indicated
points:

c)  y = x^3\: \: at \: \: (1, 1)

Answers (1)

We know that Slope of the tangent at a point on the given curve is given  by  \frac{dy}{dx}
Given the equation of the curve
y = x^3
\frac{dy}{dx}= 3x^2
at point (1,1)
\frac{dy}{dx}= 3(1)^2 = 3
Hence slope of tangent is 3
Now we know that,
slope \ of \ normal = \frac{-1}{slope \ of \ tangent} = \frac{-1}{3}
Now, equation of tangent at point (1,1) with slope = 3 is
y = mx + c\\ 1 = 1 \times 3 + c\\ c = 1 - 3 = -2
equation of tangent is
y - 3x + 2 = 0
Similarly, equation of normal at point (1,1) with slope = -1/3 is
y = mx + c
1 = \frac{-1}{3}\times 1+ c
c = \frac{4}{3}
equation of normal is
y = \frac{-1}{3}x+\frac{4}{3} \\ 3y + x = 4

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