y = sinx + 3
dy/dx = cosx
when x=π, dy/dx = cosπ = -1
so the slope = -1
also when x = π, y = sinπ + 3 = 3
so the point of contact is (π,3)
equation of tangent:
y-3 = -1(x-π)
y = -x + π+3
Find an equation of the line tangent to the graph of y=sinx+3 at X=pi.
The answer key says y=-X+Pi+3
But I got y=-X+5.
Where did the Pi even come from?
3 answers
Oh I see! Would you mind helping me with another? y=Tanx/X at X=Pi/4 find the equation of the normal to the curve.
y = tanx/x
dy/dx = (x sec^2 x - tanx)/x^2
when x = π/4 , which is 45°
dy/dx = ( (π/4)(2) - 1)/(π^2/16)
=(8π-16)/π^2
so the slope of the normal is π^2/(16-8π)
also when x = π/4, y = 1/(π/4)= 4/π
equation of normal:
y - 4/π = (π^2/(16-8π)(x - π/4)
make any cosmetic changes that you feel are needed.
dy/dx = (x sec^2 x - tanx)/x^2
when x = π/4 , which is 45°
dy/dx = ( (π/4)(2) - 1)/(π^2/16)
=(8π-16)/π^2
so the slope of the normal is π^2/(16-8π)
also when x = π/4, y = 1/(π/4)= 4/π
equation of normal:
y - 4/π = (π^2/(16-8π)(x - π/4)
make any cosmetic changes that you feel are needed.
1 answer