Equation of normal drawn to the graph of function defined as f(x) = sin x^2 / x , x not equal to zero and f zero is equal to zero at the origin is ?
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The normal line is perpendicular to the tangent line. Therefore, its slope is the negative reciprocal of the tangent line.
To find the slope of the tangent line at the point where x=1/2x=1/2, you evaluate the derivative of the equation y=2(x−1)3y=2(x−1)3when x=1/2x=1/2.
As you found, the derivative is
y′=6(x−1)2y′=6(x−1)2
Hence, when x=1/2x=1/2, the tangent line has slope
y′=6(12−1)2=6(−12)2=6(14)=32y′=6(12−1)2=6(−12)2=6(14)=32
Since the slope of the normal line is the negative reciprocal of the slope of the tangent line, the normal line has slope −2/3−2/3.
When x=1/2x=1/2,
y=2(x−1)3=2(12−1)3=2(−12)3=2(−18)=−14y=2(x−1)3=2(12−1)3=2(−12)3=2(−18)=−14
so the tangent line and normal line both pass through the point (1/2,−1/4)(1/2,−1/4).
If you use the point-slope form of the equation of a line
y−y0=m(x−x0)y−y0=m(x−x0)
you obtain
y+14=−23(x−12)y+14=−23(x−12)
x+y=0.
for the equation of the normal line.
To find the slope of the tangent line at the point where x=1/2x=1/2, you evaluate the derivative of the equation y=2(x−1)3y=2(x−1)3when x=1/2x=1/2.
As you found, the derivative is
y′=6(x−1)2y′=6(x−1)2
Hence, when x=1/2x=1/2, the tangent line has slope
y′=6(12−1)2=6(−12)2=6(14)=32y′=6(12−1)2=6(−12)2=6(14)=32
Since the slope of the normal line is the negative reciprocal of the slope of the tangent line, the normal line has slope −2/3−2/3.
When x=1/2x=1/2,
y=2(x−1)3=2(12−1)3=2(−12)3=2(−18)=−14y=2(x−1)3=2(12−1)3=2(−12)3=2(−18)=−14
so the tangent line and normal line both pass through the point (1/2,−1/4)(1/2,−1/4).
If you use the point-slope form of the equation of a line
y−y0=m(x−x0)y−y0=m(x−x0)
you obtain
y+14=−23(x−12)y+14=−23(x−12)
x+y=0.
for the equation of the normal line.
Adisha01:
ans. is x + y = 0
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Answer:
please thanks my answer
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