1. Problem 1 If f of x equals square root of x, then f of 64 equals 8
Problem 1 If f of x equals square root of x, then f of 64 equals 8. How closely does the tangent line function approximate values of f of x at x near 64? Y equals m (of tangent) times ( x minus x 1) plus y 1. M (of tangent) equals f prime of x equals 1 divided by 2 times square root of x. F prime of 64 equal to 1 divided by 16; (x 1 , y 1) equals (64, 8). Y equals f prime of 64 times (x minus 64) plus 8 equals 1 divided by 16 times (x minus 64) plus 8. Y equals 1 divided by 16 times x plus 4.

2. Problem 2 if y is equal to x to the fourth power minus 1 with x equal to negative 1 and delta x equal to d x equal to 0.01, evaluate and compare delta y and d y. Y equals x to the fourth power minus 1 Calculating delta y. Delta y equals f of (x plus delta x) minus f of x. Then, delta y equals f of (-0.99) minus fo of negative 1. Then delta y equals minus 0. So, delta y equals negative Calculating dy Y equals x to the fourth power minus 1. D y over d x equals 4 times x to the third power. D y equals 4 times x to the third power times d x. D y equals 4 times (-1) cubed times (0.01) equals negative 0.04 Note Delta x and delta y represent the unit change in the horizontal and vertical directions of the function. D x and d y represent the unit change in the horizontal and vertical directions of the tangent line to the function (linearization of the curve).

3. Problem 3 Find the differential d y of y equals 3 times x to the two-thirds power. Y equals 3 times x to the two-thirds power. D y over d x is equal to 2 times x to the negative one-third power. D y is equal to 2 times x to the negative one-third power times d x. D y is equal to (2 divided by the cube root of x) times d x

4. Problem 4 The measurement of the side of a square is found to be 12 inches, with a possible error of 1/64 inch. Use differentials to approximate the possible propagated error in computing the area of the square. Solution. A equals s squared. D of A is equal to 2 times s times d s D of A is equal to 2 times 12 inches times plus or minus one sixty fourth inch D of A is equal to plus or minus 3 eighths square inches is equal to plus or minus square inches Accuracy comparison of delta A to d A Delta A is equal to f of (s plus delta s) minus f of s Delta A equals A of minus A of 12 is approximately equal to square inches minus 144 square inches equals square inches Delta A is equal to A of minus A of 12 is approximately equal to square inches minus 144 square inches equals negative square inches

5. Problem 5. By approximately how much does the volume of a sphere with radius 5 inches change if the radius is decreased by 0.1 inches? V equals 4 divided by 3 times pi times r cubed. dV equals (dV divided by dr) times dr equals 4 times pi times r squared times dr. Dv equals 4 times pi times 5 inches squared times negative 0.1 inches. Dv equals negative ten times pi inches cubed equals negative 31.4 inches cubed.Accuracy comparison of delta v to dV delta v equals f of r plus delta r minus f of r. Delta v equals v of 4.9 minus v of 5 equivalent to minus equals negative 30.8 inches cubed.

6. Problem 6. When blowing soap bubbles, a particular bubble’s radius increases from 2 inches to 2.4 inches. Find the approximate increase in the area of the bubble. S equals 4 times pi times r squared. dS equals (dS divided by dr) times dr equals 8 times pi times r times dr. dS equals 8 times pi times 2 inches times 0.4 inches. dS equals 6.4 times pis inches squared equivalent to 20.1 inches squared. Accuracy comparison of delta s to dS. Delta S equals f of r plus delta r minus f of r. Delta s equals s of 2.4 minus s of 2 equivalent to 72.4 inches squared minus 50.3 inches squared equals 22.1 inches squared.

7. Recap. Determining the differential dy of a function: find the derivative of the function such as y equals f of u. Multiply the derivative dy / du or dy / d? By du or d of the independent variable. Obtain dy equals f prime of u times du. Using Differentials to find the tangent line approximation to a curve at the point of tangency (linearization of a curve). To find the approximation change in a function value.