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?
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
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.
To find the equation of the tangent line to the graph of y = sin(x) + 3 at x = pi, you need to consider the derivative of the function.
Now, the derivative of y = sin(x) + 3 can be found by taking the derivative of sin(x), which is cos(x), and the derivative of a constant (like 3) is zero. Therefore, the derivative of y = sin(x) + 3 is dy/dx = cos(x).
To find the slope of the tangent line at x = pi, substitute pi into the derivative dy/dx = cos(x). So, dy/dx = cos(pi) = -1.
The slope of the tangent line is -1.
Now, the equation of a straight line in slope-intercept form is given by y = mx + b, where m is the slope and b is the y-intercept.
Since the slope of the line is -1, the equation of the tangent line will be y = -x + b.
To find b, substitute the value of x = pi and y = sin(pi) + 3 into the equation. At x = pi, the y-coordinate is sin(pi) + 3 = 0 + 3 = 3.
So, the equation becomes 3 = -pi + b. Solving for b, we get b = pi + 3.
Therefore, the equation of the tangent line to the graph of y = sin(x) + 3 at x = pi is y = -x + (pi + 3), which can also be written as y = -x + pi + 3.
Hence, the correct equation is y = -x + pi + 3, and the value of pi comes from the substitution we made earlier to find the y-intercept.