14)Find the equations of the tangent and normal to the given curves at the indicated
points:
b) y = x^4 - 6x^3 + 13x^2 - 10x + 5 \: \: at \: \: (1, 3)

Answers (1)
G Gautam harsolia

We know that Slope of tangent at a point on given curve is given  by  \frac{dy}{dx}
Given equation of curve
y = x^4 - 6x^3 + 13x^2 - 10x + 5
\frac{dy}{dx}= 4x^3 - 18x^2 + 26x - 10
at point (1,3)
\frac{dy}{dx}= 4(1)^3 - 18(1)^2 + 26(1) - 10 = 2
Hence slope of tangent is 2
Now we know that,
slope \ of \ normal = \frac{-1}{slope \ of \ tangent} = \frac{-1}{2}
Now, equation of tangent at point (1,3) with slope = 2 is
y = 2x + 1
y  -2x = 1
Similarly, equation of normal at point (1,3) with slope = -1/2 is
y = mx + c
3 = \frac{-1}{2}\times 1+ c
c = \frac{7}{2}
equation of normal is
y = \frac{-1}{2}x+\frac{7}{2} \\ 2y + x = 7

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