Question

The equations of the tangent to the curve y=xsinx at the point (pi/2,pi/2) is

A) y=x -pi
B) y= pi/2
C)y= pi-2
D) y= x+pi/2
E) y= x

Answer 1

The equation of the tangent to the curve y = x sin(x) at the point (pi/2, pi/2) is y = x - pi/2.

Step-by-step explanation:

To find the equation of the tangent to the curve y = x sin(x) at the point (pi/2, pi/2), we need to find the slope of the tangent at that point and use the point-slope form of a line equation.

The derivative of y = x sin(x) is y' = sin(x) + x cos(x).

When x = pi/2, the slope of the tangent is y'(pi/2) = sin(pi/2) + (pi/2) cos(pi/2) = 1 + (pi/2) * 0 = 1.

Using the point-slope form with the point (pi/2, pi/2) and the slope 1, the equation of the tangent is y - pi/2 = 1(x - pi/2), which simplifies to y = x - pi/2.

Answer 2

The first thing you have to realize is that tangent is the slope of a curve on a given point. You can solve for the slope by finding the derivative of the given function. So:


`(d)/(dx) xsin(x)`

Next use product rule (I recommend watching videos if you're confused):


`= [sin(x)] + [x*cos(x)]`

Next substitute you x-value (π/2) into your derivative:


`= [sin( ( \pi)/(2) )] +[ (\pi)/(2)*cos( (\pi)/(2))]`

`= 1 + (\pi)/(2) *0`

`= 1`

So our slope at π/2 is 1. Next we use our slope-form and substitute our given value and solve for y-intercept (algebra-stuff)


`( (\pi)/(2)) = 1( (\pi)/(2)) + b`

`b = 0`

So we get our equation:


`y = x`

So our answer is E

Feel free to ask any questions.
Hopes this helps!