Section 11.6. Directional Derivatives We want to calculate the rate of change of at in the direction of a unit vector . This is the same as the slope of the tangent line to the trace of the surface in the plane that runs parallel to and passes through the point . !!"# !"#$ !"%&' $ !# $ ($
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Note that the and coordinates of any point on the trace lie on the line in the plane defined by The rate of change of direction of at is then with the chain rule. In particular, when per unit increase in the . We can find , we have From which we get the following. The Directional Derivative If is differentiable and is a unit vector, then the directional derivative of in the direction of is Example If , find the directional derivative of in the direction of . Solution: Use the above formula and the fact that a unit vector in the direction of an angle is . The Gradient Definition If is a function of and , then the gradient of is the vector function defined by Example Find the gradient of . In terms of the gradient the formula for the directional derivative becomes Example Find the directional derivative of at in the direction of . Functions of Three Variables Definition If is a function of , , and z, then the gradient of is the vector function defined by We have the following generalization of the directional derivative to three variables. The Directional Derivative If is differentiable and is a unit vector, then the directional derivative of in the direction of is Example Let . Find the directional derivative of at in the direction of . Maximizing the Directional Derivative Note If is differentiable and is a unit vector, then where is the angle between and . Since , Therefore, the maximum value of is and this occurs when (i.e. when is in the direction of , and the minimum value of is and this occurs when (i.e. when is in the direction of . Theorem Suppose is a differentiable function of two or more variables. Then the maximum value of is and it occurs when is in the direction of . The minimum value of is and it occurs when is in the direction of . Example. (a) Find the rate of change of at in the direction of the vector from to the point Q . (b) Find the direction of most rapid increase of at and find the maximum rate of change at . Tangent Planes to Level Surfaces !
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Suppose is a surface with equation and let be a point on . Assume is any curve on written in vector form that passes through (see the illustration above). Then for each value of t, If we differentiate both sides of this equation with respect to , we get by the chain rule, But this equation shows that the gradient vector is perpendicular to the tangent vector of any curve on that passes through . Therefore, the gradient vector is normal to the tangent plane to the surface at . If is a surface with equation , where is differentiable, and is a point on , then is a vector perpendicular to the tangent plane to the surface at . Further, the line through P and parallel to the gradient is called the normal line to at . !"#$%&'()*+),
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Example. Find the equations at the point tangent plane and normal line to the surface Solution: We have of the Hence, the gradient of at is and is perpendicular to the tangent plane at . Therefore, the tangent plane at has equation The normal line to the surface at P is parallel to and passes through . Hence, symmetric equations of the normal line are