Find the equation of the line that is tangent to the curve x(t) = t^2+1 and y(t)=t^3-1 at the point (5,7).
I get a slope of 7.5.
dy/dt = 3t^2
dx/dt = 2t
at(5,7), t=2, so dy/dx = (dy/dt)/(dx/dt) = 3t/2 = 6/2 = 3
now use that and the point-slope form to find the tangent line.
To find the equation of the line that is tangent to a curve at a given point, we need to find two pieces of information: the derivative of the curve at that point (which gives us the slope of the tangent line), and the coordinates of the point of tangency.
Let's start by finding the derivative of the curve x(t) = t^2 + 1 and y(t) = t^3 - 1. To find the derivative, we differentiate each component of the curve with respect to t:
dx/dt = 2t
dy/dt = 3t^2
Now, let's find the slope of the tangent line by evaluating the derivatives at the point (5, 7):
dx/dt = 2t
dy/dt = 3t^2
Plugging in t = 5:
dx/dt = 2(5) = 10
dy/dt = 3(5^2) = 3(25) = 75
So, the slope of the tangent line at the point (5, 7) is 10/75, which simplifies to 2/15. It appears that you made an error in calculating the slope.
To find the equation of a line given a point and a slope, we can use the point-slope form:
y - y1 = m(x - x1)
Where (x1, y1) are the coordinates of the point, and m is the slope.
Plugging in the values (x1, y1) = (5, 7) and m = 2/15, we have:
y - 7 = (2/15)(x - 5)
Now, let's simplify the equation:
y - 7 = (2/15)x - 2/3
y = (2/15)x + 49/3
So, the equation of the line tangent to the curve x(t) = t^2 + 1 and y(t) = t^3 - 1 at the point (5, 7) is y = (2/15)x + 49/3.