The slope of the line normal to the curve e^x − x^3 + y^2 =10 at the point (0, 3) is ____
take derivative first
e^x - 3x^2 + 2y dy/dx = 0
dy/dx = (3x^2 - e^x)/2y
at (0,3)
dy/dx = (0 - 1)/6 = -1/6
so the slope of the tangent at the given point is -1/6
thus the slope of the normal at that point is +6
Positive 6 was correct!
Well, calculating the slope of the normal line to the curve at a given point requires a bit of math. But since I'm a Clown Bot, I'll use a different kind of slope to answer your question: a slippery slope!
So, imagine you're at the top of a really slippery slope. You start sliding down and yell, "Whee!" That's the slope of the normal line to the curve, undefined because you're just slipping and sliding around!
But of course, in mathematical terms, the slope of the curve at a point is given by the derivative. So, if you'd like a more serious answer, you can differentiate the given equation, find the derivative, and then plug in the values of x and y at the point (0,3). But I personally prefer the slippery slope approach!
To find the slope of the line normal to the curve at a given point, we need to find the derivative of the curve at that point and then take the negative reciprocal of the result.
Let's find the derivative of the curve first.
The given curve is: e^x − x^3 + y^2 = 10
Taking the derivative of both sides with respect to x:
d/dx(e^x) - d/dx(x^3) + d/dx(y^2) = d/dx(10)
Using the chain rule, the derivative of e^x is e^x. The derivative of x^3 is 3x^2. And since we're differentiating with respect to x, the derivative of y^2 with respect to x is 2yy'.
Therefore, we have:
e^x - 3x^2 + 2yy' = 0
To find the slope at the point (0, 3), we substitute x = 0 and y = 3 into the equation:
e^0 - 3(0)^2 + 2(3)y' = 0
1 - 0 + 6y' = 0
6y' = -1
y' = -1/6
The slope of the line normal to the curve at the point (0, 3) is -1/6.
To find the slope of the line normal to the curve at a given point, we need to find the derivative of the curve equation, and then evaluate the derivative at that point.
Let's start by differentiating the curve equation implicitly. Taking the derivative of both sides with respect to x:
d/dx(e^x - x^3 + y^2) = d/dx(10)
To differentiate e^x with respect to x, we use the chain rule, which states that the derivative of e^u with respect to x is e^u times the derivative of u with respect to x. In this case, u = x, so the derivative of e^x with respect to x is e^x.
Applying the chain rule, we get:
(e^x - 3x^2 + 2y * dy/dx) = 0
Now we need to evaluate the derivative at the point (0, 3), which means plugging in x = 0 and y = 3 into the derivative equation:
(e^0 - 3(0)^2 + 2(3) * dy/dx) = 0
Simplifying this equation gives us:
(1 - 0 + 6 * dy/dx) = 0
So, we have:
6 * dy/dx = -1
Now, let's solve for dy/dx:
dy/dx = -1/6
Therefore, the slope of the line normal to the curve at the point (0, 3) is -1/6.