find an equation of the line tangent to y= sqrt(25-x^2) at the point (3,4)
i got y= x/8 + 29/8
this is wrong. but im not sure why. can you explain thank you
y=√(25-x^2)
y' = -x/√(25-x^2)
y'(3) = -3/4
so, now you have a point and a slope. The line is thus
y-4 = -3/4 (x-3)
To find the equation of the tangent line to a curve at a certain point, you need to find the derivative of the curve and evaluate it at that point.
Given the equation of the curve y = √(25 - x^2), let's first find the derivative:
Take the derivative of y with respect to x using the chain rule, as follows:
y = √(25 - x^2) = (25 - x^2)^(1/2)
Using the chain rule, the derivative of y with respect to x is:
dy/dx = (1/2)(25 - x^2)^(-1/2)(-2x)
= -x / √(25 - x^2)
Now, let's find the slope of the tangent line at the point (3, 4):
To find the slope of the tangent line, substitute x = 3 into the derivative:
dy/dx = -(3) / √(25 - (3)^2)
= -3 / √16
= -3/4
The slope of the tangent line is -3/4.
To find the equation of the tangent line, we have the point (3, 4) on the line and the slope (-3/4). We can use the point-slope form:
y - y1 = m(x - x1), where (x1, y1) is the point on the line and m is the slope.
Plugging in the values, we get:
y - 4 = (-3/4)(x - 3)
Expanding and rearranging:
y - 4 = (-3/4)x + (9/4)
y = (-3/4)x + (9/4) + 4
y = (-3/4)x + (9/4) + (16/4)
y = (-3/4)x + (25/4)
So, the equation of the tangent line to y = √(25 - x^2) at the point (3, 4) is y = (-3/4)x + (25/4), which is different from the equation you got.