# How do you find the equation of the line tangent to #y=sqrt(t-1)# at (5,2)?

The tangent line in point-slope form would be,

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To find the equation of the line tangent to the curve y = √(t - 1) at the point (5, 2), we need to find the derivative of the function and evaluate it at t = 5.

The derivative of y with respect to t can be found using the chain rule:

dy/dt = (1/2) * (t - 1)^(-1/2) * (d/dt) (t - 1)

Simplifying this expression, we get:

dy/dt = (1/2) * (t - 1)^(-1/2)

Now, we can substitute t = 5 into the derivative to find the slope of the tangent line at t = 5:

dy/dt = (1/2) * (5 - 1)^(-1/2) = (1/2) * 4^(-1/2) = 1/4

So, the slope of the tangent line at t = 5 is 1/4.

Using the point-slope form of a linear equation, we can write the equation of the tangent line:

y - 2 = (1/4)(t - 5)

Simplifying this equation, we get:

y = (1/4)t - (5/4) + 2

y = (1/4)t - (5/4) + 8/4

y = (1/4)t + 3/4

Therefore, the equation of the line tangent to y = √(t - 1) at the point (5, 2) is y = (1/4)t + 3/4.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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