140k views
3 votes
Find the slope of the normal line to y=x+cos(xy) at (0,1)

User Jesusverma
by
8.8k points

2 Answers

4 votes
Derivating the equation:


y=x+\cos(xy)\Longrightarrow y'=1+(-y\sin(xy)-xy'\sin(xy))\iff \\\\y'=1-y\sin(xy)-xy'\sin(xy)\iff y'+xy'\sin(xy)=1-y\sin(xy)\iff \\\\y'=(1-y\sin(xy))/(1+x\sin(xy))

So the slope of the tangent in the point (0,1) is:


y'=(1-y\sin(xy))/(1+x\sin(xy))\Longrightarrow y'=(1-1\sin(0\cdot1))/(1+0\sin(0\cdot1))\Longrightarrow y'=1

Then, the slope of the normal (n) line in the point (0,1) is:


n\cdot y'=-1\Longrightarrow n\cdot1=-1\iff\boxed{n=-1}
User Ezeke
by
7.6k points
7 votes

Answer with explanation:

The equation of the curve is:

y= x + Cos ( x y)

Differentiating once to get,slope of tangent

y'=1 + Sin (x y)(y + x y')

Now, slope of tangent of the curve,at (0,1) can be obtained by,substituting,x=0 and y=1 , in the equation of the slope of tangent to the curve


y'_((0,1))=1 + Sin (0 * 1)(1+0 * y'_((0,1)))\\\\ \text{as},Sin 0^(\circ)=0\\\\y'_((0,1))=1

Slope of tangent = 1

For, any curve,the two lines passing through , (0,1),that is normal line and tangent line will be perpendicular to each other,so as their slopes.

So,if two lines are perpendicular to each other

Product of their Slopes = -1

So, →Slope of tangent × Slope of Normal = -1

→ 1 ×Slope of Normal=-1

→Slope of Normal= -1

Equation of line passing through, (0,1) and having slope=-1,that is equation of normal line is

→y-1= -1×(x-0)

→ y-1=-x

x + y-1=0→→→Equation of normal line

Slope of normal line = -1

User Scott Heaberlin
by
8.4k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories