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For the following exercises, find d f d t using the chain rule and direct substitution.

f ( x , y ) = x 2 + y 2 , x = t , y = t 2

d f d t = 2 t + 4 t 3

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f ( x , y ) = x 2 + y 2 , y = t 2 , x = t

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f ( x , y ) = x y , x = 1 t , y = 1 + t

d f d t = −1

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f ( x , y ) = x y , x = e t , y = 2 e t

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f ( x , y ) = ln ( x + y ) , x = e t , y = e t

d f d t = 1

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f ( x , y ) = x 4 , x = t , y = t

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Let w ( x , y , z ) = x 2 + y 2 + z 2 , x = cos t , y = sin t , and z = e t . Express w as a function of t and find d w d t directly. Then, find d w d t using the chain rule.

d w d t = 2 e 2 t in both cases

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Let z = x 2 y , where x = t 2 and y = t 3 . Find d z d t .

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Let u = e x sin y , where x = t 2 and y = π t . Find d u d t when x = ln 2 and y = π 4 .

2 2 t + 2 π = d u d t

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For the following exercises, find d y d x using partial derivatives.

sin ( 6 x ) + tan ( 8 y ) + 5 = 0

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x 3 + y 2 x 3 = 0

d y d x = 3 x 2 + y 2 2 x y

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sin ( x + y ) + cos ( x y ) = 4

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x 2 2 x y + y 4 = 4

d y d x = y x x + 2 y 3

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x e y + y e x 2 x 2 y = 0

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x 2 / 3 + y 2 / 3 = a 2 / 3

d y d x = y x 3

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x cos ( x y ) + y cos x = 2

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e x y + y e y = 1

d y d x = y e x y x e x y + e y ( 1 + y )

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Find d z d t using the chain rule where z = 3 x 2 y 3 , x = t 4 , and y = t 2 .

d z d t = 42 t 13

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Let z = 3 cos x sin ( x y ) , x = 1 t , and y = 3 t . Find d z d t .

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Let z = e 1 x y , x = t 1 / 3 , and y = t 3 . Find d z d t .

d z d t = 10 3 t 7 / 3 × e 1 t 10 / 3

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Find d z d t by the chain rule where z = cosh 2 ( x y ) , x = 1 2 t , and y = e t .

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Let z = x y , x = 2 cos u , and y = 3 sin v . Find z u and z v .

z u = −2 sin u 3 sin v and z v = −2 cos u cos v 3 sin 2 v

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Let z = e x 2 y , where x = u v and y = 1 v . Find z u and z v .

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If z = x y e x / y , x = r cos θ , and y = r sin θ , find z r and z θ when r = 2 and θ = π 6 .

z r = 3 e 3 , z θ = ( 2 4 3 ) e 3

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Find w s if w = 4 x + y 2 + z 3 , x = e r s 2 , y = ln ( r + s t ) , and z = r s t 2 .

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If w = sin ( x y z ) , x = 1 3 t , y = e 1 t , and z = 4 t , find w t .

w t = cos ( x y z ) × y z × ( −3 ) cos ( x y z ) x z e 1 t + cos ( x y z ) x y × 4

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For the following exercises, use this information: A function f ( x , y ) is said to be homogeneous of degree n if f ( t x , t y ) = t n f ( x , y ) . For all homogeneous functions of degree n , the following equation is true: x f x + y f y = n f ( x , y ) . Show that the given function is homogeneous and verify that x f x + y f y = n f ( x , y ) .

f ( x , y ) = 3 x 2 + y 2

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f ( x , y ) = x 2 + y 2

f ( t x , t y ) = t 2 x 2 + t 2 y 2 = t 1 f ( x , y ) , f y = x 1 2 ( x 2 + y 2 ) 1 / 2 × 2 x + y 1 2 ( x 2 + y 2 ) 1 / 2 × 2 y = 1 f ( x , y )

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f ( x , y ) = x 2 y 2 y 3

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The volume of a right circular cylinder is given by V ( x , y ) = π x 2 y , where x is the radius of the cylinder and y is the cylinder height. Suppose x and y are functions of t given by x = 1 2 t and y = 1 3 t so that x and y are both increasing with time. How fast is the volume increasing when x = 2 and y = 5 ?

34 π 3

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The pressure P of a gas is related to the volume and temperature by the formula P V = k T , where temperature is expressed in kelvins. Express the pressure of the gas as a function of both V and T . Find d P d t when k = 1 , d V d t = 2 cm 3 /min, d T d t = 1 2 K/min, V = 20 cm 3 , and T = 20 ° F .

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The radius of a right circular cone is increasing at 3 cm/min whereas the height of the cone is decreasing at 2 cm/min. Find the rate of change of the volume of the cone when the radius is 13 cm and the height is 18 cm.

d V d t = 1066 π 3 cm 3 / min

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The volume of a frustum of a cone is given by the formula V = 1 3 π z ( x 2 + y 2 + x y ) , where x is the radius of the smaller circle, y is the radius of the larger circle, and z is the height of the frustum (see figure). Find the rate of change of the volume of this frustum when x = 10 in ., y = 12 in., and z = 18 in .

A conical frustum (that is, a cone with the pointy end cut off) with height x, larger radius y, and smaller radius x.
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A closed box is in the shape of a rectangular solid with dimensions x , y , and z . (Dimensions are in inches.) Suppose each dimension is changing at the rate of 0.5 in./min. Find the rate of change of the total surface area of the box when x = 2 in ., y = 3 in., and z = 1 in .

d A d t = 12 in . 2 / min

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The total resistance in a circuit that has three individual resistances represented by x , y , and z is given by the formula R ( x , y , z ) = x y z y z + x z + x y . Suppose at a given time the x resistance is 100 Ω , the y resistance is 200 Ω , and the z resistance is 300 Ω . Also, suppose the x resistance is changing at a rate of 2 Ω / min , the y resistance is changing at the rate of 1 Ω / min , and the z resistance has no change. Find the rate of change of the total resistance in this circuit at this time.

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The temperature T at a point ( x , y ) is T ( x , y ) and is measured using the Celsius scale. A fly crawls so that its position after t seconds is given by x = 1 + t and y = 2 + 1 3 t , where x and t are measured in centimeters. The temperature function satisfies T x ( 2 , 3 ) = 4 and T y ( 2 , 3 ) = 3 . How fast is the temperature increasing on the fly’s path after 3 sec?

2 ° C/sec

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The x and y components of a fluid moving in two dimensions are given by the following functions: u ( x , y ) = 2 y and v ( x , y ) = −2 x ; x 0 ; y 0 . The speed of the fluid at the point ( x , y ) is s ( x , y ) = u ( x , y ) 2 + v ( x , y ) 2 . Find s x and s y using the chain rule.

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Let u = u ( x , y , z ) , where x = x ( w , t ) , y = y ( w , t ) , z = z ( w , t ) , w = w ( r , s ) , and t = t ( r , s ) . Use a tree diagram and the chain rule to find an expression for u r .

u r = u x ( x w w r + x t t r ) + u y ( y w w r + y t t r ) + u z ( z w w r + z t t r )

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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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