How do you find an equation of the tangent line to the curve at the given point #y=sinx+3# and #x=pi#?

Question

Aurora Alvarado · Accepted Answer

#y+x=3+pi#

we use

#y-y_1=m(x-x_1)#

where #(x_1,y_1) # is a known coordinate; and #m=#gradient

#y=sinx+3#

#y(pi)=sinpi+3=0+3=3#

#(x_1,y_1) =(pi, 3)#

gradient of tangent #m=((dy)/(dx))_(x=pi)#

#y=sinx+3#

#(dy)/(dx)=cosx#

#m=((dy)/(dx))_(x=pi)=cospi=-1#

eqn tgt

#y-3=-1(x-pi)#

#y-3=-x+pi#

#y+x=3+pi#

Aurora Alvarado · Answer

undefined To find the equation of the tangent line to the curve at the given point, we need to find the slope of the tangent line and the coordinates of the point.

First, we find the derivative of the function y = sin(x) + 3. The derivative of sin(x) is cos(x), so the derivative of y with respect to x is dy/dx = cos(x).

Next, we substitute x = pi into the derivative to find the slope of the tangent line at that point. cos(pi) = -1, so the slope of the tangent line is -1.

Now, we have the slope (-1) and the point (pi, sin(pi) + 3) = (pi, 3) on the curve. We can use the point-slope form of a linear equation to find the equation of the tangent line.

The point-slope form is y - y1 = m(x - x1), where (x1, y1) is the point on the line and m is the slope. Substituting the values, we get y - 3 = -1(x - pi).

Simplifying the equation, we have y - 3 = -x + pi.

Finally, rearranging the equation, we get the equation of the tangent line to the curve y = sin(x) + 3 at x = pi as y = -x + (pi + 3).