Final answer:
To find the equation of the tangent line to s(t) = √(16 -t) when t = 0, we need to find the derivative of s(t) and evaluate it at t = 0. The derivative of s(t) is s'(t) = -1/(2√(16 - t)). When t = 0, s'(t) = -1/(2√16) = -1/8. Now we have the slope of the tangent line, which is -1/8. Using the point-slope form, the equation of the tangent line is y - √16 = -1/8(t - 0) or simplified as y = -1/8t + 2.
Step-by-step explanation:
To find the equation of the tangent line to s(t) = √(16 -t) when t = 0, we need to find the derivative of s(t) and evaluate it at t = 0.
The derivative of s(t) is s'(t) = -1/(2√(16 - t)).
When t = 0, s'(t) = -1/(2√16) = -1/8.
Now we have the slope of the tangent line, which is -1/8. Using the point-slope form, the equation of the tangent line is y - √16 = -1/8(t - 0) or simplified as y = -1/8t + 2.