🏷️ LIMITED TIME OFFER: GET 20% OFF GRADE+ YEARLY SUBSCRIPTION →

HomeClass Notes1,200,000US530,000

L24 Math 233 Lecture Notes - Lecture 18: Level Set, Directional Derivative, Implicit Function

21 views3 pages

Verified Note

6 Oct 2018

School

Department

Course

Professor

For unlimited access to Class Notes, a Class+ subscription is required.

L24 Math 233 verified notes

L24 Math 233 Lecture Notes - Lecture 17: Unit Vector, Directional Derivative, Partial Derivative

L24 Math 233 Lecture Notes - Lecture 18: Level Set, Directional Derivative, Implicit Function

L24 Math 233 Lecture Notes - Lecture 19: Level Set, Nissan L Engine, Unit Vector

Document Summary

L24 math 233 lecture 18- directional derivatives and gradient vectors. If f(x,y) is differentiable, then f(x,y) = 0 defines y = y(x) implicitly. This method can be used to solve for dx/dy does not equal 0 given dx = y dy. Write this as f(x,y) = 0 using f(x,y)= x2 + y2 1 y = 1 x2. Y = 2 dy = x dx for y, to get. X 1 x2 dy = ( x 1 x2 dx to this problem. Let u = be a unit vector (this means magnitude of u is 1) Let f = f(x,y) be differentiable at ( Definition: the directional derivative of f at ( x0 y0. <0,1> b) a, u in the u direction is. Error vanishes fast enough to ignore while calculating derivative. Cancel out h and use this approximation to calculate the directional derivative. 0 y0 * ( x0 + f. 0 y0 * h + f (x ,

Related textbook solutions

Calculus

ISBN: 9781319050733

Single Variable Calculus: Early Transcendentals

4th Edition, 2018

ISBN: 9781337687805

CALCULUS:EARLY TRANSCENDENTALS

ISBN: 9781319050740

Related Documents

L24 Math 233 Lecture Notes - Lecture 40: Nissan L Engine, Antiderivative, C More Entertainment

L24 Math 233 Lecture Notes - Lecture 23: Lagrange Multiplier, Nissan L Engine, Linear Independence

L24 Math 233 Lecture Notes - Lecture 17: Unit Vector, Directional Derivative, Partial Derivative

Related Questions

help

Using the function y = f(x) obtained in (c), find the .y coordinate corresponding to the value x = -1 / 2 Evaluate both formulas for dy / dx at this point. Why do you think this happens.? Using the explicit derivative formula dy / dx = f (x) obtained in part (c) , find all points on the graph of y = f(x) where the tangent line is horizontal. Then substitute the x and y - coordinates of each such point into the implicit dy / dx formula given in part(b). Again, why does this happen? Again using the explicit derivative formula dy / dx = f(x) from part (c). find all points on the graph of f where the tangent slope equals 6. Now substitute the x- and y- coordinates of each such point into the implicit dy / dx formula given in part (b) to confirm this slope value. It is clear that no point (x, y) for which x + 2y = 0 will appear on the graph of original equation 2x = y - x2 / x +2y. In effect, points on the line x + 2y = 0 are "cut out" of the graph of the function f Relate this observation to the phenomena encountered earlier in parts (e) and (f). Once again referring to the explicit function f(x) obtained in part (c), where does the graph of f have a vertical asymptote? What happens when you substitute this X-value into the original implicit equation (given at the beginning of this worksheet) and attempt to solve for y ? Explain. [ Verification of the implicit derivative given in part (b).] Implicitly differentiate the equation 2x = y - x2 / x = 2y, applying the quotient role on the right-hand side. Clear the denominator and expand terms, then solve for dy / dx and simplify. Note: Solutions to this problem set are t0 be established by traditional algebraic or arithmetic means. using Maple software to verify each result To download the Maple template file prepared to assist you w ith these exercises, go to https;//my,ltu edu and login, select Calculus 1 and click on Workshop. Saw this file to your laptop hard drive and exit the web Hand-generated solutions are to be fully documented In the workshop report including all steps requiring algebraic or calculus computations - merely listing the answers produce with Maple is not sufficient. Printouts of your Maple verifications are required as part of your workshop report. Consider the equation 2x = y -x2 / x+ 2y. Verify that the points .A(1, -1) and B(-2, 4/3) lie on the graph of this equation. In part (j) of this worksheet you will be asked to verify- the derivative formula by implicitly differentiating the equation given above. Evaluate this derivative at points A and B. Solve for y in the original equation, expressing y as an explicit function of x. (For later reference, we will call this function f(x).) Compute dy / dx from this expression and simplify , then evaluate this explicit version of dy / dx at both A aud B. Considering the results of part (b), are these the values you expected? Why or why not? Evaluate each of the two formulas for dy / dx (both implicit and explicit) at the point C(-1, 0). Can you explain your results?

help

Using the function y = f(x) obtained in (c), find the .y coordinate corresponding to the value x = -1 / 2 Evaluate both formulas for dy / dx at this point. Why do you think this happens.? Using the explicit derivative formula dy / dx = f (x) obtained in part (c) , find all points on the graph of y = f(x) where the tangent line is horizontal. Then substitute the x and y - coordinates of each such point into the implicit dy / dx formula given in part(b). Again, why does this happen? Again using the explicit derivative formula dy / dx = f(x) from part (c). find all points on the graph of f where the tangent slope equals 6. Now substitute the x- and y- coordinates of each such point into the implicit dy / dx formula given in part (b) to confirm this slope value. It is clear that no point (x, y) for which x + 2y = 0 will appear on the graph of original equation 2x = y - x2 / x +2y. In effect, points on the line x + 2y = 0 are "cut out" of the graph of the function f Relate this observation to the phenomena encountered earlier in parts (e) and (f). Once again referring to the explicit function f(x) obtained in part (c), where does the graph of f have a vertical asymptote? What happens when you substitute this X-value into the original implicit equation (given at the beginning of this worksheet) and attempt to solve for y ? Explain. [ Verification of the implicit derivative given in part (b).] Implicitly differentiate the equation 2x = y - x2 / x = 2y, applying the quotient role on the right-hand side. Clear the denominator and expand terms, then solve for dy / dx and simplify. Note: Solutions to this problem set are t0 be established by traditional algebraic or arithmetic means. using Maple software to verify each result To download the Maple template file prepared to assist you w ith these exercises, go to https;//my,ltu edu and login, select Calculus 1 and click on Workshop. Saw this file to your laptop hard drive and exit the web Hand-generated solutions are to be fully documented In the workshop report including all steps requiring algebraic or calculus computations - merely listing the answers produce with Maple is not sufficient. Printouts of your Maple verifications are required as part of your workshop report. Consider the equation 2x = y -x2 / x+ 2y. Verify that the points .A(1, -1) and B(-2, 4/3) lie on the graph of this equation. In part (j) of this worksheet you will be asked to verify- the derivative formula by implicitly differentiating the equation given above. Evaluate this derivative at points A and B. Solve for y in the original equation, expressing y as an explicit function of x. (For later reference, we will call this function f(x).) Compute dy / dx from this expression and simplify , then evaluate this explicit version of dy / dx at both A aud B. Considering the results of part (b), are these the values you expected? Why or why not? Evaluate each of the two formulas for dy / dx (both implicit and explicit) at the point C(-1, 0). Can you explain your results?

Show transcribed image text

raspberrychinchilla970