The rates at which the perimeter, height, area, and vertex angle of an isosceles triangle change when the legs are increasing at a given rate can be found using calculus and the properties of triangles.
The question involves an isosceles triangle with a base of 10 inches and legs that are x inches in length. To find the rates at which various measurements are changing when x is increasing at 4 inches per minute and x = 13 inches, we can use related rates from calculus and geometry.
Perimeter
The perimeter (P) of the triangle is given by P = base + 2x. The rate of change of the perimeter (dP/dt) when the legs are x inches and increasing at 4 inches per minute is dP/dt = 0 + 2(dx/dt) = 2(4) = 8 inches per minute.
Height
To find the height (h) of the triangle, we can use the Pythagorean theorem in one-half of the isosceles triangle. Expressing h in terms of x, we get h = √(x² - (base/2)²). The rate of change of height (dh/dt) is found by implicit differentiation. dh/dt can be calculated by substituting x = 13 inches and dx/dt = 4 in/min into the derived formula.
Area
The area (A) of a triangle is A = 1/2 × base × height. We can find the rate of change of area (dA/dt) by differentiating A concerning time and using the known values for height, dP/dt, and dh/dt.
Vertex Angle
The rate at which the vertex angle (ϴ) is changing can be found by differentiating the equation for the cosine of ϴ, which relates the vertex angle to the legs and base of the triangle, and then solving for dϴ/dt.