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Directional Derivatives via Gradient - Calculus 3

  1. The function z=f(x,y)z = f(x,y) has partial derivatives fx(2,3)=4f_x(2,3) = 4 and fy(2,3)=−1f_y(2,3) = -1. A curve CC passes through the point (2,3)(2,3) and has tangent vector t⃗=3i⃗−4j⃗\vec{t} = 3\vec{i} - 4\vec{j} at that point. What is the directional derivative of ff along the curve CC at the point (2,3)(2,3)?
    Answer: 165\frac{16}{5} The gradient at (2,3)(2,3) is ∇f(2,3)=(4,−1)\nabla f(2,3) = (4, -1). The tangent vector t⃗=(3,−4)\vec{t} = (3, -4) has magnitude ∣∣t⃗∣∣=9+16=5||\vec{t}|| = \sqrt{9 + 16} = 5. The unit tangent vector is u⃗=(35,−45)\vec{u} = (\frac{3}{5}, -\frac{4}{5}). The directional derivative along the curve is ∇f⋅u⃗=(4)(35)+(−1)(−45)=125+45=165\nabla f \cdot \vec{u} = (4)(\frac{3}{5}) + (-1)(-\frac{4}{5}) = \frac{12}{5} + \frac{4}{5} = \frac{16}{5}. Choice B ignores the direction and just gives fxf_x. Choice C makes a calculation error. Choice D has the wrong sign, possibly from incorrectly handling the tangent vector direction.
  2. Consider the function g(x,y)=ln⁡(x2+y2)g(x,y) = \ln(x^2 + y^2) at the point (3,4)(3,4). If the maximum rate of increase of gg at this point is kk, what is the directional derivative of gg in the direction making an angle of π6\frac{\pi}{6} with the direction of maximum increase?
    Answer: 2325\frac{2\sqrt{3}}{25} First find ∇g=(2xx2+y2,2yx2+y2)\nabla g = (\frac{2x}{x^2+y^2}, \frac{2y}{x^2+y^2}). At (3,4)(3,4): ∇g(3,4)=(625,825)\nabla g(3,4) = (\frac{6}{25}, \frac{8}{25}). The magnitude is ∣∣∇g∣∣=36625+64625=1025=25||\nabla g|| = \sqrt{\frac{36}{625} + \frac{64}{625}} = \frac{10}{25} = \frac{2}{5}. The unit vector in direction of max increase is u⃗1=(35,45)\vec{u}_1 = (\frac{3}{5}, \frac{4}{5}). A direction making angle π6\frac{\pi}{6} with this has directional derivative ∣∣∇g∣∣cos⁡(π6)=25⋅32=2325||\nabla g|| \cos(\frac{\pi}{6}) = \frac{2}{5} \cdot \frac{\sqrt{3}}{2} = \frac{2\sqrt{3}}{25}. Choice A uses wrong denominator. Choice C has coefficient error. Choice D uses wrong base calculation.
  3. The temperature at point (x,y)(x,y) is given by T(x,y)=100−x2−2y2T(x,y) = 100 - x^2 - 2y^2. A particle at point (3,2)(3,2) moves in the direction of the vector w⃗=−i⃗+2j⃗\vec{w} = -\vec{i} + 2\vec{j}. If the particle moves at speed 3 units per second, what is the rate of change of temperature experienced by the particle?
    Answer: −65-6\sqrt{5} degrees per second The gradient is ∇T=(−2x,−4y)\nabla T = (-2x, -4y). At (3,2)(3,2): ∇T(3,2)=(−6,−8)\nabla T(3,2) = (-6, -8). The unit vector in direction of w⃗=(−1,2)\vec{w} = (-1, 2) is u⃗=(−1,2)5=(−15,25)\vec{u} = \frac{(-1, 2)}{\sqrt{5}} = (-\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}}). The directional derivative is ∇T⋅u⃗=(−6)(−15)+(−8)(25)=65−165=−105=−25\nabla T \cdot \vec{u} = (-6)(-\frac{1}{\sqrt{5}}) + (-8)(\frac{2}{\sqrt{5}}) = \frac{6}{\sqrt{5}} - \frac{16}{\sqrt{5}} = -\frac{10}{\sqrt{5}} = -2\sqrt{5}. Since the particle moves at speed 3, the rate of change is 3×(−25)=−653 \times (-2\sqrt{5}) = -6\sqrt{5}. Choice B has wrong sign. Choice C incorrectly multiplies by 9 instead of 3. Choice D forgets to normalize the direction vector.

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