In Exercise 69 and 70, find curl
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Chapter 15 Solutions
Calculus
- Find the unit vectors that are parallel to the tangent of the polynomial y= x^3 +2x + 1 at the point (1,4)arrow_forwardWhat is dyldx at the point (1,2) of xy^2 + 2xy = 8 ?"arrow_forwardVECTOR DIFFERENTIATION:If A = t(^2)i - tj + (2t+1)k and B = (2t-3) i + j - tk, obtain the following at t=1:a) d/dt(A dot B)b) d/dt(A x B)c) d/dt |A + B| d) d/dt (A x dB/dt)arrow_forward
- Find a unit vector tangent to the curve of intersection of -x^2 - 2y^3 = z - 3 and 25/x2 - 4y - 3z^2 = - x + 6 at the point (1,1,0).arrow_forwardGood morning, could you please help me with the following demonstration: Prove that the set of periodic functions of a scalar variable is not a vector space.Examples: sin (x), sin (2x), sin^(x)arrow_forwardDetermine order and degree * d'y + cos dr? d'y dr? 2 O 5 undefine- Not lineararrow_forward
- Suppose given a function f(*, y, 2) of three variables, and a point (*0, y0, 20) ER. Explain in your own words how to use the gradient vectors of f(*, y, 2) to obtain an equation for the tangent plane to the surface f(*, 3, 2) = cin R' at the point (*0, 30, 20), where c = f(*0, yo, z0).arrow_forwardConsider this scalar-valued function: f(x,y)=y/ 2x^2+y^2 . After, find the gradient of f and find the directional derivative of f at the point (1,3) in the direction of v=(1,−2) please show detailsarrow_forwardConsider the function f : R 2 → R: f(x, y) = ax^2 + by^2 , where a and b are real constants. (a) Assume a, b > 0. Sketch a contour plot for the function f, including the level set at 1. (b) Assume a > 0 and b < 0. Sketch a contour plot, including the level set at 1. (c) Describe the level set at 1 when a = 0, b ̸= 0 and when a ̸= 0, b = 0.arrow_forward
- Find a vector equation for the curve of intersectionof the surfaces x^2 + y^2 = 4 andz = xy.arrow_forwardShow that t, e^t, and sin(t) are linearly independent.arrow_forwardEvaluate fx, fy, fz at the given point: a) f (x, y z) = x³yz² at the point (1, 2, 3) b) f (x, y, z) = x² - 2xy + 3yxz² at the point (3, 1, -2) c) f (x, y, z) = 2xe ^ y - 3ye ^ -x at the point (3, 1, -2)arrow_forward
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