Thread: Writing the equation of a tangent line

1. Writing the equation of a tangent line

I'm reviewing for the AP exam and would love it if someone could walk me through this problem- I know you find the derivative of the curve at the given point... but I get lost from there
The slope of a function f at any point (x,y) is y/(2x)^2. The point (2,1) is on the graph of f.
A. Write an equation of the tangent line to the graph f at x=2.
B. Use the tangent line in part (a) to approximate f(2.5).
C. Solve the differential equation dy/dx= y/(2x)^2 with the initial condition f(2)=1.
D. Use the solution in part c and find f(2.5)

2. Re: Writing the equation of a tangent line

Originally Posted by qubez
I'm reviewing for the AP exam and would love it if someone could walk me through this problem- I know you find the derivative of the curve at the given point... but I get lost from there
The slope of a function f at any point (x,y) is y/(2x)^2. The point (2,1) is on the graph of f.
A. Write an equation of the tangent line to the graph f at x=2.
B. Use the tangent line in part (a) to approximate f(2.5).
C. Solve the differential equation dy/dx= y/(2x)^2 with the initial condition f(2)=1.
D. Use the solution in part c and find f(2.5)
A. point $\displaystyle (2,1)$ , use $\displaystyle \frac{dy}{dx}$ to evaluate the slope at the given point ...

slope = $\displaystyle \frac{1}{(2 \cdot 2)^2} = \frac{1}{16}$

write the tangent line equation using the point-slope form ...

$\displaystyle y - 1 = \frac{1}{16}(x - 2)$

B. use the tangent line to approximate the function value at x = 2.5 ...

$\displaystyle f(x) \approx y = \frac{1}{16}(x - 2) + 1$

$\displaystyle f(2.5) \approx y = \frac{1}{16}(2.5 - 2) + 1 = \frac{33}{32}$

C. separate variables ...

$\displaystyle \frac{dy}{y} = \frac{dx}{4x^2}$

integrate ...

$\displaystyle \ln|y| = -\frac{1}{4x} + C_1$

change to an exponential to isolate y ...

$\displaystyle y = C_2 e^{-\frac{1}{4x}}$

utilize the given initial condition to solve for the constant of integration ...

$\displaystyle 1 = C_2 e^{-\frac{1}{8}}$

$\displaystyle C_2 = e^{\frac{1}{8}}$

simplify ...

$\displaystyle y = f(x) = e^{\frac{x-2}{8x}}$

D. $\displaystyle f(2.5) = e^{\frac{1}{8}}$