MAT 21C Lecture Notes - Lecture 22: Partial Derivative, Product Rule, Implicit Function

Indefinite Integrals, with substitution 6. Evaluate the following indefinite integrals. These require substitution. Show the substitution. For example, for f(x)=f8x cosh (4x2)dx f'( The substitution is u=4x2 . The derivative is a 8xdr . Substitute the u and the du into the integral. In this case, replace the 8xdx with du. Do not use an intermediate step with both old and new variables in the integral. x)=8xcosh(4x2), then f(u)-I cosh u du=sinh u+ C Reverse the substitution: ,f(x)=sinh 4x4 C 0.5 | | sin-111-cos-ldt u=1-cos- 2

EXAMPLE 5 Find the area of the largest rectangle that can be inscribed in a semicircle of radius r SOLUTION 1 Let's take the semicircle to be the upper half of the circle x2 + y2 = with center the origin. Then the word inscribed means that the rectangle has two vertices on the semicircle and two vertices on the x-axis as shown in the top figure. r,y) 2x Let (x, y) be the vertex that lies in the first quadrant. Then the rectangle has sides of lengths 2x and y, so its area is A = r a To eliminate y we use the fact that (x, y) lies on the circle x2 + y2 = r2 and so y = Thus A = The domain of this function is 0 sxs r. Its derivative is rsin θ rcos θ which is 0 when 2x2-, that is, x- (since x 0). This value of x gives a maximum value of A since A(0) = 0 and Video Example A(r) = o. Therefore the area of the largest inscribed rectangle is SOLUTION 2 A simpler solution is possible if we think of using an angle as a variable. Let θ be the angle shown in the bottom figure. Then the area of the rectangle is A(9)-(2r cos(8)(rsin(8))-r2(2 sin(8) cos(8)) = r2 sin(28) π/2. So A(8) has a maximum value of r2 and it occurs when we know that sin(26) has a maximum value of 1 and it occurs when 2θ θ = π/4. Notice that this trigonometric solution doesn't involve differentiation. In fact, we didnt need to use calculus at all

Search Textbook Solutior x 526408 Q2 Notes Ask Your Teach EXAMPLE 5 Find the area of the largest rectangle that can be inscribed in a s emicircle of radlusr SOLUTION 1 Let's take the semicircle to be the upper half of the circle x+y Then the word inscribed means that the rectangle has two vertices on the axis as shown in the top figure. with center the semicircle and two vertices on the x- Let (x, y) be the vertex that les in the first quadrant. Then the rectangle has sides of lengths 2x and y, so its area is A = To eliminate y we use the fact that (x, y) lles on the circle x2+ y2 2 and so y . Thus The domain of this function is 0 s x s r. Its derivative is 2x22-2x2 which is 0 when 2x2=r2, that is, x = (since x 2 0). This value of x gives a maximum value of A since A(O)-0 and A(r) 0. Therefore the area of the largest inscribed rectangle is SOLUTION 2 A simpler solution is possible if we think of using an angle as a variable. Let 8 be the angle shown in the bottom figure. Then the area of the rectangle is A(0)-(2r cos(θ))(r sin(θ))-r2(2 sin(θ) cos(θ)) = r2 sin(28). We know that sin(20) has a maximum value of 1 and it occurs when 2θ = π/2. So A(9) has a maximum value of r2 and it occurs when θ = π/4. Notice that this trigonometric solution doesn't involve differentiation. In fact, we didn't need to use calculus at all. ^ d, 7:30 PM 11/26/2017