Solve the problem.
-Find an equation for the tangent to the curve at $P$.

A) $y = \sqrt { 2 } x + 1 + \sqrt { 2 }$
B) $y = - \sqrt { 2 } x + \frac { \pi \sqrt { 2 } } { 4 } - 1 - \sqrt { 2 }$
C) $y = \sqrt { 2 } x - \frac { \pi \sqrt { 2 } } { 4 } + 1 + \sqrt { 2 }$
D) $y = - \sqrt { 2 } x - 1 - \sqrt { 2 }$

Question

Solve the problem.
-Find an equation for the tangent to the curve at $P$.

A) $y = \sqrt { 2 } x + 1 + \sqrt { 2 }$ 
B) $y = - \sqrt { 2 } x + \frac { \pi \sqrt { 2 } } { 4 } - 1 - \sqrt { 2 }$ 
C) $y = \sqrt { 2 } x - \frac { \pi \sqrt { 2 } } { 4 } + 1 + \sqrt { 2 }$ 
D) $y = - \sqrt { 2 } x - 1 - \sqrt { 2 }$

Quizplus · Accepted Answer

To find the equation of the tangent, we need two things: the slope of the tangent and a point on the tangent.

First, let's find the slope of the tangent at point $P$. We can do this by taking the derivative of the curve at $P$:

$$\frac{d}{dx}(\sin x + \cos x) = \cos x - \sin x$$

Plugging in $x = \frac{\pi}{4}$ (the x-coordinate of point $P$), we get:

$$\cos \frac{\pi}{4} - \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0$$

So the slope of the tangent at point $P$ is 0.

Next, let's find a point on the tangent. We know that point $P$ is on the tangent, so we can use that. The coordinates of point $P$ are $\left(\frac{\pi}{4},\sqrt{2}\right)$.

Using the point-slope form of a line, we can write the equation of the tangent as:

$$y - \sqrt{2} = 0\cdot \left(x - \frac{\pi}{4}\right)$$

Simplifying, we get:

$$y = \sqrt{2}x - \frac{\pi\sqrt{2}}{4} + \sqrt{2} + 1$$

Which matches answer choice C.

Solve the Problem PPP A) y=2x+1+2y = \sqrt { 2 } x + 1 + \sqrt { 2 }y=2​x+1+2​

Solve the Problem $P$

A) $y = \sqrt { 2 } x + 1 + \sqrt { 2 }$