Question

The equation to the normal to the curve \[y = \sin x\] at \[(0,0)\] is
A. \[x = 0\]
B. \[y = 0\]
C. \[x + y = 0\]
D. \[x - y = 0\]

Answer 1

Hint: The equation of the line passing through \[({x_1},{y_1})\] and having slope \[m\] is \[(y - {y_1}) = m(x - {x_1})\] . Since slope of a normal is given by \[m = - \dfrac{{dx}}{{dy}}\] . If we differentiate \[y = \sin x\] with respect to x we get \[\dfrac{{dy}}{{dx}}\] and taking the reciprocal of the obtained solution we get \[\dfrac{{dx}}{{dy}}\] . Putting \[({x_1},{y_1}) = (0,0)\] we obtained the slope. Substituting these in \[(y - {y_1}) = m(x - {x_1})\] we get the required equation.

Complete step-by-step answer:
To get the required equation we need to find the slope.
We know that slope of a normal is \[ \Rightarrow m = - \dfrac{{dx}}{{dy}}\] ---- (1)
Now to find \[\dfrac{{dx}}{{dy}}\] .
Differentiate \[y = \sin x\] with respect to x.
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = \cos x\] (We know differentiation of sine is cosine)
Taking the reciprocal of this we get,
 \[ \Rightarrow \dfrac{{dx}}{{dy}} = \dfrac{1}{{\cos x}}\]
Substituting in equation (1) we get,
 \[ \Rightarrow m = - \dfrac{1}{{\cos x}}\]
We have \[({x_1},{y_1}) = (0,0)\] that is \[x = 0\]
 \[ \Rightarrow m = - \dfrac{1}{{\cos 0}}\]
Since, \[\cos 0 = 1\] we get,
 \[ \Rightarrow m = - 1\]
Slope of the normal is -1.
We know the equation of the line passing through \[({x_1},{y_1})\] and having slope \[m\] is \[(y - {y_1}) = m(x - {x_1})\] .
We have \[({x_1},{y_1}) = (0,0)\] and \[m = - 1\] then substituting we get,
 \[ \Rightarrow (y - 0) = - 1(x - 0)\]
 \[ \Rightarrow y = - x\]
 \[ \Rightarrow y + x = 0\]
 \[ \Rightarrow x + y = 0\]
The equation to the normal to the curve \[y = \sin x\] at \[(0,0)\] is \[x + y = 0\]
So, the correct answer is “Option C”.

Note: We know the slope of a tangent is \[\dfrac{{dy}}{{dx}}\] . If they ask the same question like, find the equation of tangent to the curve. Then follow the same procedure that we did in above but slope m will change while remaining steps are the same. Don’t get confused with the slope of normal and the slope of tangent. Both are different (normal are perpendicular to the tangent and tangent is a line that touches the curve only at one point).