The normal (perpendicular) line to the curve y=sqrt(8-x^2) at (-2,2) has slope
A. -2
B. 1/2
C. -1/2
D. 1
E. -1
I got A
y = √(8-x^2)
y' = -x/√(8-x^2)
y'(-2) = 2/2 = 1
so, the normal has slope -1
Well, well, well. Looks like we have a normal line on our hands. Let's get this straight, shall we? Pun intended.
To find the normal line to a curve, we need the slope of the tangent line at the point of interest first. We can find that by taking the derivative of the function y = √(8 - x^2).
If I put on my math cape and do some calculations, I'll get the derivative as: dy/dx = -x/√(8 - x^2).
Now, to find the slope at the point (-2, 2), we plug in x = -2 into our derivative expression:
dy/dx = -(-2)/√(8 - (-2)^2) = 2/√(8 - 4) = 2/√4 = 2/2 = 1.
So, it seems like the slope of the tangent line at (-2, 2) is 1. But what about the slope of the normal line?
Well, the normal line will be perpendicular to the tangent line, so the slopes will be negative reciprocals of each other. In other words, if the tangent's slope is 1, then the normal line's slope will be -1.
So, dear friend, it seems like the answer is E. -1. Just remember, when it comes to slopes, opposites attract!
To find the slope of the normal line to the curve at a given point, we need to find the derivative of the curve at that point.
The given curve is y = sqrt(8 - x^2). To find the derivative, we can use the chain rule. Let's first rewrite the equation as y = (8 - x^2)^(1/2).
Using the chain rule, we have:
dy/dx = (1/2)(8 - x^2)^(-1/2)(-2x)
Now, let's evaluate this derivative at the point (-2, 2). Substitute x = -2 into the derivative equation:
dy/dx = (1/2)(8 - (-2)^2)^(-1/2)(-2)(-2)
= (1/2)(8 - 4)^(-1/2)(-2)(-2)
= (1/2)(4)^(-1/2)(-2)(-2)
= (1/2)(2)^(-1)(-2)(-2)
= (1/2)(1)(-2)(-2)
= (-2)
The slope of the normal line is equal to the negative reciprocal of the derivative at the given point. Therefore, the slope of the normal line to the curve at (-2, 2) is -2.
So, the correct answer is A.
To find the slope of the normal line to the curve at a given point, we need to find the derivative of the curve and then determine the negative reciprocal of that derivative.
Step 1: Find the derivative of the curve y = sqrt(8 - x^2).
The given function represents a portion of a circle centered at the origin. We can rewrite it as y = (8 - x^2)^(1/2). To find the derivative, we can use the chain rule since the square root function is involved:
dy/dx = (1/2) * (8 - x^2)^(-1/2) * (-2x)
= -x / sqrt(8 - x^2)
Step 2: Find the negative reciprocal of the derivative.
The negative reciprocal of a number is when you flip the fraction and change the sign. Therefore, the negative reciprocal of -x / sqrt(8 - x^2) is sqrt(8 - x^2) / x.
Step 3: Substitute the x-coordinate of the given point (-2,2) into the negative reciprocal.
Substituting x = -2 into the negative reciprocal, we get sqrt(8 - (-2)^2) / (-2) = sqrt(8 - 4) / (-2) = sqrt(4) / (-2) = 2 / (-2) = -1.
Therefore, the slope of the normal line to the curve at (-2,2) is -1.
Since you got A, which is -2, it seems there may have been an error in your calculation. Please check your work again.