Math Problem Statement

find an equation of the tangent line to f(t) at a=-1

Find an equation of the tangent line to f(t) = √(3t + 8) at a = -1.

Solution

To find the equation of the tangent line to the function f(t)=3t+8f(t) = \sqrt{3t + 8} at a=1a = -1, we can follow these steps:

  1. Find the point of tangency: This is f(1)f(-1).
    f(1)=3(1)+8=5.f(-1) = \sqrt{3(-1) + 8} = \sqrt{5}. So, the point of tangency is (1,5)(-1, \sqrt{5}).

  2. Find the slope of the tangent line: This is given by f(1)f'(-1), which is already calculated in the problem as 3510\frac{3\sqrt{5}}{10}.

  3. Use the point-slope form of the equation of a line: The point-slope form is: yy1=m(tt1),y - y_1 = m(t - t_1), where mm is the slope and (t1,y1)(t_1, y_1) is the point of tangency. Substituting m=3510m = \frac{3\sqrt{5}}{10}, t1=1t_1 = -1, and y1=5y_1 = \sqrt{5}, we get: y5=3510(t+1).y - \sqrt{5} = \frac{3\sqrt{5}}{10}(t + 1).

  4. Simplify the equation: y=5+3510(t+1).y = \sqrt{5} + \frac{3\sqrt{5}}{10}(t + 1).

This is the equation of the tangent line in exact form.

Would you like a more detailed explanation of any part of the process?

Related Questions:

  1. How do we find the derivative f(t)f'(t) using the definition of the derivative?
  2. Why is the point-slope form used for tangent lines?
  3. What is the difference between finding a secant line and a tangent line?
  4. How does the chain rule apply when differentiating functions like 3t+8\sqrt{3t + 8}?
  5. What does the slope of the tangent line represent geometrically?

Tip: The point-slope form is very useful for writing the equation of a tangent line because it directly uses the slope and a known point on the curve.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Tangent Lines

Formulas

f'(t) = lim (t -> a) [(f(t) - f(a)) / (t - a)]
Point-slope form: y - y1 = m(t - t1)

Theorems

Definition of the derivative
Point-slope formula for tangent lines

Suitable Grade Level

Grades 11-12 (Advanced Calculus)