find equations for the tangent
and the normal at P (1n2, 2k)
on the curve with equation
y=ke^x, where k is a constant.
(b) given that y=sin^-1x, show
that (1-x^2)d^2y/dx^2-xdy/ dx=0
(a) no idea what 1n2 means
(b)
y = arcsin(x)
y' = 1/√(1-x^2)
y" = x/(1-x^2)3/2
so,
(1-x^2)y" - xy'
= x/√(1-x^2) - x/√(1-x^2)
= 0
(a) Ah. I think that's ln2 (not 1n2)
y' = ke^x
y'(ln2) = 2k
So, now we have a point and a slope. Back to algebra I:
tangent: y-2k = 2k(x-ln2)
normal: y-2k = -1/2k (x-ln2)
To find the equations for the tangent and the normal at point P(1,2) on the curve with equation y = ke^x, where k is a constant, we need to find the derivative of the curve and evaluate it at the point P.
First, let's find the derivative of y with respect to x:
dy/dx = d/dx (ke^x)
To differentiate ke^x with respect to x, we can use the chain rule. The derivative of e^x is e^x, and the derivative of k is 0 since it is a constant. Thus, the derivative of ke^x with respect to x is ke^x.
Now that we have the derivative dy/dx, we can evaluate it at the point P(1,2). Plugging in x=1 into dy/dx, we get:
dy/dx = k * e^1 = k * e
So, the slope of the tangent line at P is k * e.
The equation for the tangent line can be written using the point-slope form of a line: y - y1 = m(x - x1), where (x1, y1) is the point P and m is the slope.
Using P(1,2) and k * e as the slope, we get:
y - 2 = k * e * (x - 1)
This is the equation of the tangent line at point P on the curve y = ke^x.
To find the equation for the normal line, we need to find the negative reciprocal of the slope of the tangent line. Let's call it m_normal.
m_normal = -1 / (k * e)
Now, we can use the point-slope form again to get the equation for the normal line:
y - 2 = (-1 / (k * e)) * (x - 1)
This is the equation for the normal line at point P on the curve y = ke^x.
Moving on to part (b) of the question, given that y = arcsin(x), we need to show that (1 - x^2) * d^2y/dx^2 - x * dy/dx = 0.
To find dy/dx, we differentiate y = arcsin(x) with respect to x:
dy/dx = d/dx (arcsin(x))
Taking the derivative of arcsin(x) requires the use of inverse trigonometric derivatives. The derivative of arcsin(x) with respect to x is 1 / sqrt(1 - x^2).
Now, let's find d^2y/dx^2. Taking the derivative again, we have:
d^2y/dx^2 = d/dx (1 / sqrt(1 - x^2))
To differentiate 1 / sqrt(1 - x^2), we use the chain rule. The derivative of 1 / sqrt(1 - x^2) is (d/dx(1) * sqrt(1 - x^2) - 1 * d/dx(sqrt(1 - x^2))) / (sqrt(1 - x^2))^2.
Simplifying this expression, we get (0 * sqrt(1 - x^2) - 1 * (-2x)) / (1 - x^2).
Now, we can substitute dy/dx and d^2y/dx^2 into the equation provided:
(1 - x^2) * ((0 * sqrt(1 - x^2) - 1 * (-2x)) / (1 - x^2)) - x * (1 / sqrt(1 - x^2)) = 0
Simplifying this equation, we can see that the term (1 - x^2) cancels out:
0 - 2x + x / sqrt(1 - x^2) = 0
Combining like terms, we have:
-x + x / sqrt(1 - x^2) = 0
Multiplying through by sqrt(1 - x^2), we get:
-sqrt(1 - x^2) + x = 0
Now, let's add sqrt(1 - x^2) to both sides of the equation:
x = sqrt(1 - x^2)
This equation is equivalent to the original equation, so we have shown that (1 - x^2) * d^2y/dx^2 - x * dy/dx = 0.